Ministry of Education and Science of the Republic of Kazakhstan

«Almaty University of Energy and Communications»

Nonprofit Joint Stock Company

G.K. Vassilina

THE PROBABILITY THEORY

IN EXAMPLES AND TASKS

Manual for high school

Almaty

AUES

2019

UDC 519.2(075.8)

LBS 22.1я73

V 30

Reviewer:

Doctor of Physical and Mathematical Sciences, Professor,

Chief Researcher of the Department of Differential Equations

of the Institute of Mathematics and Mathematical Modeling

A.T. Asanova,

PhD, Assistant Professor of the Suleyman Demirel University

B.K. Zhakhaev

Ph.D., Associate Professor of the Department

of Mathematics and Mathematical Modeling, AUEC

R.E. Kim

Vassilina G.K.

V 30 Probability theory in examples and tasks: Manual for high school. – Almaty:

Almaty University of Energy and Communications, 2019. – 94 p.: tables – 34,

illustrations – 10, bibilography – 10.

ISBN 978-601-332-698-6

The textbook for high school «Probability theory in examples and tasks» is

intended for students of speciality 6В01009 – Mathematics, 6В01010 – Physics,

6В01011 – Computer science. It`s compiled in accordance with the program of

discipline.

UDC 519.2(075.8)

LBS 22.1я73

ISBN 978-601-332-698-6

© AUEC, 2019

G.K. Vassilina, 2019.

3

Introduction

Probability theory provides a mathematical foundation to concepts such as

«probability», «information», «belief», «uncertainty», «randomness», «variability»,

«chance» and «risk». Probability theory is important to empirical science because it

gives them a rational framework to make inferences and test hypotheses based on

uncertain empirical data.

The probability theory gives mathematical model for description of such a

random phenomena in objective reality. Since many real processes are subject to

random influences, it is important for experts, who devoted themselves to the

natural, technical, as well as social sciences, to know bases of this theory.

1 Elements of combinatorics

Combinatorics - the branch of mathematics that deals with collections of

objects that satisfy specified criteria (e.g., counting arrangements, permutations, and

combinations).

For example, combinatorics would answer the question «How many different

ways can you arrange a 10-song playlist if you have 45 songs to choose from?»

Combinations - the branch of combinatorics where changing the order of the

objects does not create a new scenario.

For example, the question «How many teams of 9 baseball players can a

manager assemble from a roster of 18 players?» is a combinations question since

changing the order in which the chosen player does not create a new arrangement.

Factorials. Solving permutations and combinations problems using a formula

requires the use of factorials. The k factorial function (symbol: k!) is the product of

all positive integers less than or equal to k. Examples:

- k factorial = 123...)2()1(! −−= kkkk ;

- 5 factorial, denoted 5! And 5! 5 4 3 2 1 120;= =

- 7 factorial, denoted 7! And 7! 7 6 5 4 3 2 1 5 040;= =

- 1! = 1.

Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no

numbers together gets us 1, but it helps simplify a lot of equations.

Permutations - the branch of combinatorics where changing the order of the

objects creates a new scenario. For example, the question «How many different

ways can a baseball pitcher who throws 6 unique pitches (e.g., curveball, fastball,

changeup, etc.) throw the next 3 pitches?» is a permutations question because

changing the throwing order creates a new arrangement (e.g., throwing fastball,

sinker, changeup is different than throwing sinker, changeup, fastball).

There are basically two types of permutation:

- repetition is allowed: such as the lock above. It could be "333";

4

- no repetition: for example the first three people in a running race. You can't

be first and second.

Permutations with Repetition. These are the easiest to calculate. When a thing

has n different types. We have n choices each time!

For example: choosing 3 of those things, the permutations are: nnn (n is

multiplied 3 times).

More generally: choosing r of something that has n different types, the

permutations are:

... nnn (r times).

In other words, there are n possibilities for the first choice, then there are n

possibilities for the second choice, and so on, multiplying each time.

Which is easier to write down using an exponent of r:

rnnnn = ... (r times).

1.1 Cast a coin once. The complete space of simple events is { , }, 2.H T n = 1.2 If we toss а coin twice, we саn observe 1 of 4 outcomes: (Heads, Heads),

(Heads, Tails), (Tails, Heads), (Tails, Tails). In this case we could use the complete

space contains four simple events: { , , , },HH HT TH TT

22 4.n = =

1.3 If we cast coin thee times, then:

H: HHH, HHT, HTH, HTT,

T: THH, THT, TTH, TTT.

Then the complete space contains eight simple events:

{ , , , , , , , },HHH HHT HTH HTT THH THT TTH TTT

32 8.n = =

Thus, if we cast a coin k times, then 1 2{ ... , }, 2 .kk ka a a a H or T n= =

1.4 In the lock above, there are 10 numbers to choose from

(0,1,2,3,4,5,6,7,8,9) and we choose 3 of them 310 10 10 10 1000 = = (3 times)

permutations.

So, the formula is simply: nr, where n is the number of things to choose from,

and we choose r of them (repetition is allowed, order matters)

1.5 How many ternary strings of length 4 have zero ones?

5

Solution. If there are no ones then we can only use symbols 0 and 2 so there

are 2 possibilities for each of 4 positions so the answer is 42 2 2 2 2 16 = = .

1.6 How many ternary strings of length 4 have exactly one?

Solution. Ternary strings have symbols 0, 1, and 2. If there is exactly one 1,

then there are 3 positions the one can be in and 222 ways to fill the other 3 blanks

with a 0 or a 2. So the answer is 3 2 2 2 24 = .

Permutations without Repetition. In this case, we have to reduce the number

of available choices each time.

For example, what order could 16 pool balls be in? After choosing, say,

number «14» we can't choose it again. So, our first choice has 16 possibilities, and

our next choice has 15 possibilities, then 14, 13, etc. And the total permutations are

16 15 14 ... 2 1 20 922 789 888 000 = .

But maybe we don't want to choose them all, just 3 of them, so that is only

3603141516 = .

In other words, there are 3 360 different ways that 3 pool balls could be

arranged out of 16 balls.

Without repetition our choices get reduced each time. But how do we write

that mathematically? Answer: we use the «factorial function».

So, when we want to select all of the billiard balls the permutations are

16! 20 922 789 888 000= .

But when we want to select just 3 we don't want to multiply after 14. How do

we do that? There is a neat trick: we divide by 13!

16 15 14 13 12 ... 2 1

16 15 14 3 360.13 12 ... 2 1

= =

So: 16!

16 15 14.13!

=

The formula is written:

!

( )!

n

n r−,

where n is the number of things to choose from, and we choose r of them (No

repetition, order matters).

For example, our «order of 3 out of 16 pool balls example» is:

16! 16! 20 922 789 888 000

3 360,(16 3)! 13! 6 227 020 800

= = =−

which is just the same as: 3603141516 = .

6

1.7 How many ways can first and second place be awarded to 10 people?

10! 10! 3 628 800

90(10 2)! 8! 40 320

= = =−

which is just the same as: 10 × 9 = 90.

Notation. Instead of writing the whole formula, people use different notations

such as these:

.)!(

!),(

rn

nPrnP r

n−

==

For example, .90910!8

!10

)!210(

!10)2,10( 2

10===

−== PP

1.8 If there are six students and six desks in an art class, how many different

seating arrangements are there?

Solution. There are six spots to be filled:

- the first student who enters the class has six choices (or possible seats);

- the second student has five choices (or possible seats);

- the third student has four choices (or possible seats);

- the fourth student has three choices (or possible seats);

- the fifth student has two choices (or possible seats);

- the sixth student has no choice (only one seat left).

Thus, there are 720!6123456 == seating arrangements.

Solving with the equation: n=6=number of possible seats to choose from k =

6 = number of seats that will be chosen:

.720!6!0

!6

)!66(

!666 ===

−=P

1.9 How many ways can you arrange 5 people on a ferris wheel with 6 seats?

Solution. Answer: 5! Because it is a circular permutation of essentially 6

things - 5 people and one empty seat and there are (n-1)! circular permutations of n

things. Here n = 6. If there were 4 people and 2 empty seats the answer would not

be 5! any longer since the 4 people are different, but the 2 empty seats are

indistinguishable.

For example, let a, b, c and d be the 4 people and e1 and e2 be the empty seats.

There is no real difference between

21 eedcba and 12 eedcba ,

but if you gave the answer (6-1)! = 5! you would count these identical

permutations twice. The answer for this situation would be:

7

5! 5!

5 4 3 60(5 2)! 3!

= = =−

.

1.10 In a race with 30 runners where 8 trophies will be given to the top 8

runners (the trophies are distinct: first place, second place, etc), how many ways can

this be done?

Solution: This is a permutation problem since the trophies are distinct. Think

of the trophies as being 8 positions. The answer is P(30, 8) - the number of ways to

arrange 30 things taken 8 at a time - which is:

.2324252627282930!22

!30

)!830(

!30830 ==

−=P

You can also think of drawing 8 blanks representing 8 trophies and multiply

the number of possibilities for each blank: 282930 etc.

1.11 How many ways can you do the above problem if a certain person, Aset,

must be one of the top 3 winners?

Solution. There are 3 ways to put Aset in one of the top 3 positions. Then

there are 23...2829 ways to put 7 other people in the 7 other winning positions.

So the answer is: 323...2829 .

1.12 How many ways can you arrange 16 people into 4 rows of 4 desks each?

Solution. In this problem the desks and rows are considered distinct. The

answer is 16!. It doesn't matter how the desks are arranged. You could number them

1 through 16 and the problem becomes a simple permutation of 16 things.

1.13 How many ways can you pair up 8 boys and 8 girls?

Solution. Answer: 8!. If sounds like there are 2 sets of 8 things to consider

permuting, but really we are only permuting one set of 8 things. Think of the boys

as in a fixed order:

boy 1 boy 2 ... boy 8.

Then the girls can be arranged in 8! ways. Each arrangement corresponds to

one pairing with the boys: first girl in the arrangement with boy 1, second girl in the

arrangment with boy 2, etc.

1.14 How many ways can one arrange the seven letters of the word

SYSTEM?

Solution. The two S’s are indistinguishable so we find the arrangements of

the four letters Y,T,E,M taken from a set of size six and let the two S’s occupy the

remaining slots. Hence, we have:

.3603456!2

!646 ===P

8

Another way to think of this problem is to momentarily distinguish the two

S’s as 1S and 2S . In this way there are 6! permutations of the six letters. However,

there are 2!=2 permutations of the 1S and 2S so we must divide the 6! permutations

by 2! getting the same result.

Combinations. There are also two types of combinations (remember the order

does not matter now):

- repetition is allowed: such as coins in your pocket (5,5,5,10,10);

- no repetition: such as lottery numbers (2, 14, 15, 27, 30, 33).

Combinations with Repetition. Actually, these are the hardest to explain, so

we will come back to this later.

Combinations without Repetition. This is how lotteries work. The numbers

are drawn one at a time, and if we have the lucky numbers (no matter what order)

we win!

The easiest way to explain it is to:

- assume that the order does matter (ie permutations);

- then alter it so the order does not matter.

Going back to our pool ball example, let's say we just want to know which 3

pool balls are chosen, not the order.

We already know that 3 out of 16 gave us 3360 permutations. But many of

those are the same to us now, because we don't care what order!

For example, let us say balls 1, 2 and 3 are chosen. These are the possibilites:

Order does matter Order doesn't matter

1 2 3

1 3 2

2 1 3

2 3 1

3 1 2

3 2 1

1 2 3.

So, the permutations will have 6 times as many possibilities. In fact there is

an easy way to work out how many ways «1 2 3» could be placed in order, and we

have already talked about it. The answer is: 6123!3 == . For example, 4 things can

be placed in 241234!4 == different ways, try it for yourself!

So we adjust our permutations formula to reduce it by how many ways the

objects could be in order (because we aren't interested in their order any more):

.

As this formula is so important, it is often just written in big parentheses like

this:

9

where n is the number of things to choose from, and we choose r of them (no

repetition, order doesn't matter).

It is often called «n choose r» (such as «16 choose 3»). And is also known as

the Binomial Coefficient.

Notation. As well as the «big parentheses», people also use these notations:

!

( , ) .!( )!

rn

nС n r С

r n r= =

−

So, our pool ball example (now without order) is:

.560321

161514

!13!3

!16

)!316(!3

!16)3,16( 3

16=

==

−==ÑÑ

1.15 If there are 2 intern positions open at a firm with 5 applicants, how many

different combinations of applicants could be hired?

Plugging the information from the problem into the equation:

.1021

45

!3!2

!5

)!25(!2

!525 =

==

−=Ñ

The equation tells us that if we have five applicants and are choosing two

individuals, there are 10 different ways to do this.

1.16 How many different committees of 4 students can be chosen from a

group of 15?

Solution. There are 415Ñ possible combinations of 4 students from a set of 15:

415

15! 15 14 13 121 365.

4!11! 4 3 2 1С

= = =

There are 1 365 different committees.

1.17 The combinations (or subsets) of size 2 that can be selected from the 3

letters in the word cat are {ca}, {ct}, {at}.

1.18 Find the combinations (i.e. subsets) of size 2,1=r and 3 taken from the

set {a,b,c}.

Solution. It is a simple matter to enumerate the combinations which are listed

Table 1.

10

Table 1

r = 1 {a} {b} {c}

r = 2 {a,b} {a,c} {b,c}

r = 3 {a,b,c}

Note that the 8 subsets of {a,b,c} with the exception of the empty set ∅ are

listed. If we wanted a subset of size 0 we would include the empty set.

Note that combinations are simply sets, so {a,b} is the same as the

combination {b,a}. Note too how this contrasts with permutations where the

permutation ab is not the same as ba.

1.19 How many ways can 10 players choose up sides to play five-on-five in a

game of basketball?

Solution. As in many combinatorial problems, there is more than one way to

carry out the counting. Perhaps the simplest is to determine the number of ways one

fixed player can select his or her four teammates from the 9 other players. In other

words, the number of subsets of size four taken from a set of size 9, or «nine choose

four», which:

49

9! 9 8 7 6126

4!5! 4 3 2 1С

= = =

ways.

1.20 Poker Hands. In poker, five cards are dealt from a deck of 52 cards. The

types of poker hands (and examples) are:

1) Royal Flush: Royal Flush: Royal Flush: A, K, Q, J, 10 of same suit (10♠,

S♠, J♠, Q♠, K♠, A♠).

2) Straight Flush: Five cards of same suit in sequence (4,5,6,7,8);.

3) 4 of a Kind: Four cards of the same rank (7, 7, 7, 7).

4) Full House: Three of a kind plus a pair (7♥, 7♣, 7♠, K♥, K♦).

5) Flush: Five cards of the same suit (3♥, 7♥, 10♥, Q♥, A♥).

6) Straight: Five cards in sequence (5♣, 6♦, 7♦, 8♠,9♦).

7) 3 of a Kind: Three cards of the same rank (J♥,J♠,J♦).

8) 2 Pairs: Two pairs of different rank (5♣, 5♥, 9♦, 9♠).

9) 1 Pair: Two cards of the same rank (A♣, A♥).

Find:

a) the total number of poker hands;

b) the number of 4-or-a-kinds;

c) the number of full houses;

d) the number of 3-of-a-kind hands.

Solution:

11

a) total hands: Each hand represents a subset of size 5 from a set of size 52.

Hence the total number of hands a player can receive is “52 choose 5” or:

552

52! 52 51 50 49 482 598 960

5! 47! 5 4 3 2 1С

= = =

total hands;

b) 4-of-a-kind: There are 13113 =Ñ ways to select the “quads” and 481

48 =Ñ

choices for the remaining card. Using the multiplication principle, we get:

1 148 13 48 13 6 244С С = =

four of a find hands;

c) full House: There are 13113 =Ñ choices for the rank of the triple, and 121

12 =Ñ

choices for the rank of the pair. There are also 434 =Ñ ways to choose the triple from

a card of a given rank, and 224 =Ñ ways to choose the pair from four cards of the

other rank. Hence, the multiplication principle gives us:

744364121324

34

112

113 == ÑÑÑÑ

full house hands;

d) three of a Kind: There are 13113 =Ñ choices for the rank of the triple, and

66212 =Ñ (ways) 12 choices for the rank of the remaining 2 cards. There are also

434 =Ñ choices for the triple of the given rank, and 41

4 =Ñ for each of the remaining

two cards. Using the multiplication principle, we have:

9125414

212

113 = ÑÑÑ

3-of-a-kind hands.

1.21 (Going to the Movies). Three boys and two girls are going to a movie.

How many ways can they sit next to each other under the following conditions: a)

Neither boy sits next to each other, b) The two girls sit next to each other.

Solution:

a) the only way they can sit is boy-girl-boy-girl-boy. But the boys can be

permuted 3!=6 ways and the girls 2! = 2 ways, so the total number of arrangements

is 3!2! = 12;

b) first think of the two girls are a «single girl» so you have 4 persons, 3 boys

and 1 girl. Hence there are 4! = 24 ways to permute the two girls among the 3 boys.

But, for each of these arrangements, we can permute the two girls 2! = 2 ways, and

so the total number of arrangements is 4!2! = 48.

1.22 Combinations with Repetition. Let us say there are five flavors of ice

cream: banana, chocolate, lemon, strawberry and vanilla.

12

We can have three scoops. How many variations will there be? Let's use

letters for the flavors: {b, c, l, s, v}. Example selections include:

- {c, c, c} (3 scoops of chocolate);

- {b, l, v} (one each of banana, lemon and vanilla);

- {b, v, v} (one of banana, two of vanilla).

And just to be clear: there are n = 5 things to choose from and we choose r =

3 of them. Order does not matter, and we can repeat!

Now, we show a special technique that lets our work it out. Think about the

ice cream being in boxes, we could say «move past the first box, then take 3 scoops,

then move along 3 more boxes to the end» and we will have 3 scoops of chocolate!

So it is like we are ordering a robot to get our ice cream, but it doesn't change

anything, we still get what we want.

We can write this down as (arrow means move, circle means

scoop).

In fact the three examples above can be written like this:

{c, c, c} (3 scoops of chocolate): {b, l, v} (one each of banana, lemon

and vanilla)

{b, v, v} (one of banana, two of

vanilla)

So instead of worrying about different flavors, we have a simpler question:

«How many different ways can we arrange arrows and circles?»

Notice that there are always 3 circles (3 scoops of ice cream) and 4 arrows

(we need to move 4 times to go from the 1st to 5th container).

So (being general here) there are r + (n−1) positions, and we want to choose

r of them to have circles.

This is like saying «we have r + (n−1) pool balls and want to choose r of

them». In other words it is now like the pool balls question, but with slightly

changed numbers. And we can write it like this:

.)!1(!

)!1(1

−

−+=−+

nr

nrÑ r

nr

Where n is the number of things to choose from, and we choose r of them

(Repetition allowed, order doesn't matter).

Interestingly, we can look at the arrows instead of the circles, and say "we

have r+(n−1) positions and want to choose (n−1)of them to have arrows", and the

answer is the same:

.)!1(!

)!1(111

−

−+== −

−+−+nr

nrCÑ n

nrr

nr

So, what about our example, what is the answer?

13

.35321

765

!4!3

!7

)!15(!3

)!153(=

==

−

−+

There are 35 ways of having 3 scoops from five flavors of ice cream.

In the box there are N balls, exactly M of which are white. Let n balls are

taken out from the box one by one without returning ( )n М balls. Then the

probability that among these removed n balls there will be k white is:

).,...,1,0(),(, nkC

CCknP

nN

knMN

kM

MN ==−−

A random variable is called distributed hypergeometrically, if it accepts

possible values 0,1,...,n with probabilities ),(, knP MN defined by the formula (1).

The numbers N, M, n are the distribution parameters.

1.23 How many ways can you put 10 indistinguishable desks into 3 grad

student offices (assuming each office can hold up to 10 desks!)?

Solution. r =10, n=3 so the answer is C(10+3-1,10)=C(12,10).

1.24 How many ways can you choose 12 doughnuts from 20 different types

(assuming there are at least 12 of each type)?

Solution. Imagine the 12 doughnuts you are choosing are indistinguishable

dots. The 20 different types of doughnut are like 20 distinguishable boxes. So this is

the same type of question and the answer is C(20+12-1, 20). This is another

common formulation of the same kind of problem. Below we summarize the three

ways these kinds of problems can be presented:

Identical Objects. Sometimes identical objects are involved in the

arrangements. If the identical objects are switched, the arrangement remains

unchanged. Thus, these two different arrangements are actually the same. The

formula to determine the number of arrangements possible when identical objects

are involved is as follows:

.

Where n stands for the total number of objects, and the ki stands for the

number of the ith type of object.

1.25 How many different unique combinations of letters can be created by

rearranging the letters in MATHEMATICS?

1.26. If a class of 10 students has five men, how many ways can the men and

women be arranged in a circle so that no two men sit next to each other?

1.27 Example 1: How many ways there are to place 7 balls 3 in boxes?

1.28 Eggs are drawn from a basket with 20 eggs and placed into a bowl. How

many different bowls with 6 eggs could exist?

14

1.29 7 balls are drawn from a basket with 50 balls and placed in order of

selection on the table. How many different ball orderings exist?

1.30 How many ways there are to select 10 coins from a pile with 1c, 5c, 10c

and 25c coins (consider all coins of the same value equivalent)?

1.31 Suppose we have to form a number consisting of three digits using the

digits 1,2,3,4. To form this number the digits have to be arranged. Different

numbers will get formed depending upon the order in which we arrange the digits.

1.32 If an experiment can be performed in «n» ways, & another experiment

can be performed in «m» ways then either of the two experiments can be performed

in (m+n) ways. This rule can be extended to any finite number of experiments.

1.33 Suppose there are 3 doors in a room, 2 on one side and 1 on other side.

A man want to go out from the room. Obviously he has «3» options for it. He can

come out by door «A» or door «B» or door «C»:

1.34 How many different signals can be made by 5 flags from 8-flags of

different colours? Answer 6 720.

1.35 How many words can be made by using the letters of the word

«SIMPLETON» taken all at a time? Answer 362880.

1.36 How many ways can the letters of the word «Pre-University» be

arranged?

1.37 A child has 3 pocket and 4 coins. How many ways can he put the coins

in his pocket. Answer 81.

1.38 How many ways can a cricket-eleven be chosen out of 15 players? If (i)

A particular player is always chosen; (ii) A particular is never chosen. Answer: (i) 1

365; (ii) 364.

1.39 Find the number of different choices that can be made from 3 apples, 4

bananas and 5 mangoes, if at least one fruit is to be chosen. Answer: 119.

1.40 How many ways 5 balls can be selected from «12» identical red balls?

Answer 119.

1.41 Example. How many ways can you choose 4 groups of 4 people from 16

people, assuming the groups are distinct? Answer 44

48

412

416 ÑÑÑÑ .

1.42 How many ways can you arrange 30 people on a ferris wheel with 30

seats? Answer 29! since it is a circular permutation.

1.43 How many permutations are there of the word «repetition»? Answer

10!/(2!2!2!).

15

1.44 Pigeon Hole Principle. Apply the pigeonhole to prove that at least two

people in New York City have the same number of hairs on their head. Hint: You

may want to make a few assumptions regarding the population of NYC and the

maximum number of hairs on the human head.

1.45 Subsets of a Set. Give another proof that the number of subsets of a set

of size n is n2 . Hint: Assign to each binary number of at most n digits in the

following way. Assign a 1 if the corresponding element in the set is selected to be in

the subset, otherwise a zero.

1.46 Derangements. A derangement is a permutation in which none of the

elements remain in their natural order. For example the only derangements of

(1,2,3) are (3,1,2) and (2,3,1). Hence we write !3=2. Nicolas Bernoulli proved that

the number of derangements of a set of size n is:

=

−=

n

k

k

knn

1 !

)1(!! .

How many derangements are there for the members (1,2,3,4)? Enumerate

them.

1.47 Counting Functions. How many functions are there from A={a,b,c} to

B={0,1,2}? Write them down and draw the graphs for a few of them.

1.48 One-to-One Functions. How many one-to-one functions are there from

A={a,b,c} to the set of binary numbers B={0,1}?

1.49 Onto Functions. How many onto functions are there from A={a,b,c} to

the set of binary numbers B={0,1}?

1.50 Going to the Movies. Four girls and four boys are going to a movie.

How many ways can they be seated if no two girls sit next to each other?

1.51 Baseball Season. A baseball league consists of 9 teams. How many

games will be played over the course of a year if each team plays every other team

exactly 20 times?

1.52 How many 2-element sets are there in the set }1001:{ nNx such that

the sum of two elements is even?

1.53 Picky People. How many ways can 8 people sit next to each other at a

movie if a certain 2 of them refuse to sit next to each other?

1.54 One Committee. How many ways can the Snail Darter Society, which

has 25 members, elect an executive committee of 2 members?

1.55 Two Committees. How many ways can the Snail Darter Society, which

has 25 members, elect an executive committee of 2 members and an entertainment

committee of 4 members if no member of the society can serve on both

committees?

1.56 Three Committees. How many ways can the Snail Darter Society, which

has 25 members, elect an executive committee of 2 members, an entertainment

committee of 3 members, and a welcoming committee of 2 members if no member

of the society can serve on more than one committees?

16

1.57 Serving on More than One Committee. How many ways can the Snail

Darter Society, which has 25 members, elect an executive committee of 2 members,

an entertainment committee of 3 members, and a welcoming committee of 2

members if members can serve on more than one committee?

1.58 Counting Softball Teams. A college softball team is taking 25 players on

a road trip. The traveling squad consists of 3 catchers, 6 pitchers, 8 infielders, and 6

outfielders. Assuming each player can only play her own position, how many

different teams can the coach put on the field?

1.59 Permutations as Groups. The 3! = 6 permutations of set {1,2,3} can be

illustrated by the 2×3 arrays:

,213

321,

132

321,

321

321

=

=

=

=

=

=

123

321,

312

321,

231

321 ,

where the bottom row of each array shows how the top row is permuted.

Carrying out one permutation followed by another defines a multiplication of the

permutations. For instance = means we do permutation β first then

permutation δ second, which yields the permutation ε (check it yourself). Compute

the following:

a) ; b) 2 ; c) 3 3 α; d) .

After you get the hang of multiplying permutations, make a 6×6

multiplication table of all products. This table describes what in group theory is

called the symmetric group of order 3, denoted by 3S .

1.60 Lotteries played 16 tickets. Among them there are 6 tickets to win.

Bought 5 tickets. Find the probability that the two will be the winning ticket.

1.61 A batch of 20 sewing machines has 2 faulty machines. Choose 5 random

test machines. Find the probability that it will be one fault and four serviceable

machines among them.

1.62 Among the group of 15 students (which has 8 girls) 5 tickets to the

concert are played. Find the probability that it will be 3 girls among the ticket

holders.

1.63 The batch of the 12 pieces has 7 standard. Find the probability that it

would be exactly 5 standard among the 6 pieces taken at random.

1.64 From an urn containing 9 white and 3 black balls, 4 balls are randomly

removed. Find the probability that 3 of them will be white.

1.65 In the national team of the Faculty of chess there are 12 people,

including 7 first-raters. Randomly choose 5 members of the team. Find the

probability that 3 of them will be the first-raters.

17

1.66 There are 20 details in box, 10 of them are painted. Picker randomly

extracted 3 details. Find the probability that it will be painted; a) the 2 details b) all

details.

1.67 there are 6 men and 4 women in the shop. According to personnel

numbers 7 people are randomly selected. Find the probability that it will be 2

women among the selected individuals.

1.68 Lotteries played 15 tickets. Among them there are 6 tickets to win. Find

the probability that 2 tickets will be winning from 5 tickets randomly selected.

1.69 An urn has 7 white and 3 black balls. 6 balls are randomly taken out.

What is the probability that it will be 4 white balls and 2 black balls among them?

1.70 There are 5 simple and 10 colored pencils in the box. 6 pencils are

randomly extracted out of the box. Find the probability that 2 of them are simple.

1.71 Three cards are removed at random from a deck of 36 cards. Find the

probability that it will be 2 aces among them.

1.72 An urn has 16 balls, including 10 red, the other are black. Find the

probability that it will be 3 red balls among 4 balls randomly taken out.

1.73 The party of 10 items has 7 standards. Find the probability that among

the 7 would be taken at random items exactly 5 standards.

1.74 One Committee. How many ways can the Snail Darter Society, which

has 25 members, elect an executive committee of 2 members? Members can serve

on more than one committee?

2 Algebra of events

Simple events. The probability is a very important branch of mathematics

which in addition to statistics plays a vital role in maths, science, chemistry,

psychology, geology, mining etc. Probability is a science of study of possibilities of

happening event. In probability, we usually calculate the chances of an event to be

happened.

Definition. The outcome of the experiment is called the simple event.

For example, casting a coin there is heads falling, i.e., H – «heads» is the

elementary event, T – «tails» is the elementary event. For example, infinite Sample

Space. Flip a coin until heads appears for the first time:

,...},,,,{ TTTTHTTTHTTNTHHS = .

Often we are not interested in individual outcomes, but in events. An event is

a subset of a sample space.

2.1 With respect to S1, describe the event B of rolling a total of 7 with the

two dice:

)}1,6(),2,5(),3,4(),4,3(),5,2(),6,1{(=B .

The event B can be represented graphically (figure 1):

18

Dice 2

6 * ■ ■ ■ ■ ■

5 ■ * ■ ■ ■ ■

4 ■ ■ * ■ ■ ■

3 ■ ■ ■ * ■ ■

2 ■ ■ ■ ■ * ■

1 ■ ■ ■ ■ ■ *

Dice 1

1 2 3 4 5 6

Figure 1

Often we are interested in combinations of two or more events. It can be

represented using set theoretic operations. Assume a sample space S and two events

A and B are given:

- complement A or cA : all elements of S that are not in A;

- subset A⊆ B: all elements of A are also elements of B;

- union A∪B or A+B: all elements of S that are in A or B;

- intersection A∩B or AB: all elements of S that are in A and B.

There are 3 basic set operations:

Union: the union of two sets А and В is another set that includes аll elements

of А and аll elements of В. We represent the union operator with this symbol or

symbol +.

For example, if А={1,3,5} and В={2,3,4}, then }5,4,3,2,1{=+= BABA .

More generally:

{ : }A B A or B = .

If other words, the set АВ is the set of elements with the property that they

either belong to the set А or to the set В.

Intersection: the intersection of two sets А and В is another set С such that аll

elements in С belong to А and to В. The intersection operator is symbolized as .

For example, if А={1,3,5} and В={2,3,4} then }3{== ABBA .

More generally:

}:{ BandABA = .

Complementation: the complement of а set А with respect to а universal set

is the set of аll elements of which do not belong to А. The complement of А is

represented as A or cA .

For example, if the universal set is {l,2,3,4,5,6} then the complement of

{l,3,5} is {2,4,6}.

19

More generally:

{ : }A and A = .

Empty set: the empty set is а set with nо e1ements. We represent the пиll set

with the symbol . For any set A:

;A A A+= = .

For actions with random events (sums, products, complements), formulas for

elementary theory of sets hold:

1) Commutativity:

,A B B A AB BA+ = + = .

2) Associativity:

)()( CBACBA ++=++ ; )()( BCACBA = .

3) Distributivity:

)()()( BCACCBA +=+ ; ))(()( CBCACBA ++=+ .

4) De Morgan's formulas:

, ,

, .

A B A B AB A B

A A AA

+ = = +

+ = =

2.2 Prove equalities BABA =+ , BABA += .

Solution. The sum of BA+ means that it is happened at least one of the

events A and B. Therefore, opposite event BA + is equivalent to that neither event A,

no event B appeared. But the same thing means the product BA . We proved the

first equality. The sum and product have equal rights, hence the second equality is

also true.

2.3 Show that AAA =+ , AAA= .

Solution. These equalities are obvious. Since the appearance of at least one of

two identical events and their combination is equivalent to the appearance of the

same event.

2.4 Prove that =+++++ ))(())(( BABABABA .

Solution. Opening the brackets on the left and using the result of the Example

75 and the properties of the algebra of events:

20

A A+ = , AA = and =+ AA , =AA ,

where is a reliable event, is an impossible event. We obtain:

( )( ) ( )( )A B A B A B A B+ + + + + =

A AB AB BB A A B AB BB= + + + + + + + =

( ) ( )A A B B A A B B

= + + + + + + + =

.A A A A A A= + + + = + =

2.5 Two events A and B are given from the field S. Prove that

BAABAAAA +=== 321 ,, form a complete group of incompatible events.

Solution. According to the Example 75 we have ABBAA =+=3 .

Since:

====== BBAAAABBAAAA 3121 ,

==== ABBABABAAA 32 .

It means, that 321 ,, AAA are incompatible events. Besides:

1 2 3 ( )

,

A A A A AB AB A A B B

A A A A

+ + = + + = + + =

= + = + =

i.e. the events 321 ,, AAA are form a complete group.

2.6 Prove, that:

a) ABBABABA =+++ ))()(( ;

b) =++++ ))()()(( BABABABA ;

c) =+++ ABABA ))(( ;

d) BAABBABA +=++ ))(( .

3 Direct calculation of probabilities

Probability of simple events. Terminology for probability theory:

- experiment: process of observation or measurement; e.g., coin flip;

- outcome: result obtained through an experiment; e.g., coin shows tails;

21

- sample space: set of all possible outcomes of an experiment; e.g., sample

space for coin flip: S={H,T}. Sample spaces can be finite or infinite.

Probability is the chance or likelihood that an event will happen. It is the ratio

of the number of favorable outcomes to the number of possible outcomes. The

probability of an event A is equal to the sum of the probabilities of the sample

events contained in A. Events are denoted by capital letters A, B, C, etc

A simple event is an event that consists of exactly one outcome or we can say

that, a simple event is the event of a single outcome. A probability value is assigned

to each simple event using either relative frequency or personal probabilities. The

notation P (A) is used to denote the probability of the event A to occur.

In order to measure probabilities, mathematicians have devised the following

formula for finding the probability of an event:

outcomespossibleofnumbertotalThe

occurcanAeventwaysofnumberTheAP =)( .

The probability P(A) of event A is equal to the sum of the probabilities of the

outcomes favorable to it:

1 2( ) ... .nP A p p p= + + +

The particular case n

ppp n

1...21 ==== leads to the formula:

( ) .m

P An

=

Formula (2) expresses the so-called classical definition of probability

according to which the probability of some event A is equal to the ratio of the

number m of outcomes favorable to A to the number n of all «equally likely»

outcomes.

Axioms of Probability. The probability of an event A is denoted by P(A).

1. The probability of an event is a nonnegative real number: P(A)≥0 for any

A⊆S.

2. P(S)=1.

3. If ,...,, 321 AAA , is a set of mutually exclusive events of S, then:

...)()()(...)( 321321 +++= APAPAPAAAP

Definition. Event A, which must occur during the experiment, is called the

certain event and P(A)=1.

22

Definition. Event B, which must not occur during the experiment, is called the

impossible event and P(B)=0.

0)( =P for any sample space S.

Definition. Event, which occurs during the experiment, as A fails, is called an

opposite event. An opposite event is denoted by А .

We have 1)()( =+ APAP or )(1)( APAP −= . The probability of sum of events

equals the sum of their probabilities:

1 2

1

( ... ) ( )n

n i

i

P A A A P A=

+ + + = .

Below you could see some examples of the probability:

3.1 There are 4 blue marbles, 5 red marbles, 1 green marbles, and 2 black

marbles in a bag. Suppose you select one marble at random. Find probability.

Solution. Sample space: 12. There are 12 marbles total (4+5+1+2=12):

Probability .(

)

The number of outcomes conducive to

Total Possible Outcomes Sample S

ou

pa

r e ent

ce

v=

P(black)=2/12=1/6 There are 2 black marbles in the bag 12 is

your sample space.

P(blue)=4/12=1/3 There are 4 blue marbles in the bag 12 is

your sample space.

P(blue or black) =

=6/12=1/2

4 blue + 2 black = 6. 12 is your sample

space.

P(not green)=11/12 There is 1 green, so 12 – 1 = 11 that aren`t

green 12 is your sample space.

P(not purple)=1 I will definitely select a marble that is not

purple because there are no purple marbles in the bag.

Whenever the chance of something occurring is

definite, the probability is 1.

3.2 Toss two fair coins and record the outcome. Find the probability of

finding the exactly one head in the two tosses.

Solution. Sample space for the tossing two coins = {HH, HT, TH, TT}. Here

the letters H and T means a head or a tail respectively. Since the sum of the four

simple events must be 1, each must have probability ¼, then P(A)=1/4+1/4=1/2.

23

3.3 A single 6-sided die is rolled. What is the probability of each outcome?

What is the probability of rolling an even number? of rolling an odd number?

P(7)=?

Solution. The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.

Probabilities:

1 1

(1) ;6

numberof ways to rollP

total number of sides= =

2 1(2) ;

6

numberof ways to rollP

total number of sides= =

3 1(3) ;

6

numberof ways to rollP

total number of sides= =

4 1(4) ;

6

numberof ways to rollP

total number of sides= =

5 1(5) ;

6

numberof ways to rollP

total number of sides= =

6 1(6) ;

6

numberof ways to rollP

total number of sides= =

3 1( ) ;

6 2

numberof ways to roll an even numberP even

total number of sides= = =

3 1( ) .

6 2

numberof ways to roll an odd numberP odd

total number of sides= = =

0)7( =P . This is impossible because the die does not contain a number 7.

Whenever the probability is impossible, the answer is 0.

Task 3.3 illustrates the difference between an outcome and an event. A single

outcome of this experiment is rolling a 1, or rolling a 2, or rolling a 3, etc. Rolling

an even number (2, 4 or 6) is an event, and rolling an odd number (1, 3 or 5) is also

an event.

In Task 3.3 the probability of each outcome is always the same. In Task 82,

the probability of rolling each number on the die is always one sixth. Let's look at

an experiment in which the outcomes are not equally likely

3.4 A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a

single marble is chosen at random from the jar. What is the probability of choosing

a red marble? a green marble? a blue marble? a yellow marble?

24

Outcomes: The possible outcomes of this experiment are red, green, blue and

yellow. Total number of marbles:

6 + 5 + 8 + 3 = 22;

6 3

( ) ;22 11

numberof ways to choose redP red

total number of marbles= = =

5( ) ;

22

numberof ways to choose greenP green

total numberof marbles= =

8 4( ) ;

22 11

numberof ways to choose blueP blue

total number of marbles= = =

3( ) .

22

numberof ways to choose yellowP yellow

total numberof marbles= =

The outcomes in this experiment are not equally likely to occur. You are

more likely to choose a blue marble than any other color. You are least likely to

choose a yellow marble.

Summary. The probability of an event is the measure of the chance that the

event will occur as a result of an experiment. The probability of an event A is the

number of ways event A can occur divided by the total number of possible

outcomes.

We will use a number of examples related to a standard bridge deck of 52

cards. Such a deck is made up as shown in the following figure:

3.5 What is the probability of drawing out an ace from a well shuffled deck of

cards?

25

Solution. There are 52 outcomes (the number of cards in the deck), they are

equally likely (from a well-shuffled deck), and there are 4 ways to obtain an ace, so

we have P=4/52, or 1/13.

3.6 In the tossing of two dice, each of the 36 possible outcomes can be

designated by ),( ji , where i is the number of pips that comes up on the first dice and

j, the number on the second. The outcomes are assumed to be equally likely. To the

event A, “the sum of the pips is 4,” three outcomes are favorable: (1;3); (2;2); (3;1).

Consequently, P(A)=3/36=1/12.

The actual finding of probability of some random event by purely theoretical

way is often impossible. In these cases, it is necessary to make a sufficient number

of tests and take the relative frequency of the event as an approximate value of

probability.

So, for example, there is no mathematical or biological theory that allows to

calculate a priori the probability P(A) of a random event "the birth of twins". To

determine P(A), it is necessary to use the statistics of a large number of births and to

calculate how often this event has been occurred. Then the corresponding frequency

can be approximately taken as the probability P(A).

The definition of P(A) by frequency is sometimes called the «statistical

definition» of probability. But here, however, we speak not about the definition of

probability, but about its evaluation.

Let the experiment be repeated n times and the quantity of occurrence of

interesting for us event be counted. Suppose that it occurred m times. The ratio:

*( ) ( )m

P A W An

= =

is called the relative frequency (or briefly - the frequency) of a random event A in n

experiments. Here 1)(0 * AP .

The formulation of axioms is based on the following properties of frequency,

which can be considered as experimental facts:

1) For large n, the frequency )(* AP oscillates less and less around a certain

value.

2) 1)(* =UP , where U is reliable.

3) )()()( *** BPAPBAP += , If A and B are incompatible.

In the classical case, combinatorial formulas are often used. For example, in

order to calculate the number of all possible elementary events.

3.7 Lotto game: guess «k numbers from n». For example, bingo 6 out of 49.

What is the probability of getting the main prize? Indicate the number k. There is

one favorable event in which the main prize of the game is won. The number of all

elementary events is equal to the number of possible samples of k numbers from n

26

without counts of order and without repeats, i.e. it is equal to knÑ . Thus, the

probability of a main winning in a bingo is:

.81698313

11649

=C

In the box there are N balls, exactly M of which are white. Let n balls be

taken out from the box one by one without returning ( )n М balls. Then the

probability that among these removed n balls there will be k white is:

).,...,1,0(),(, nkC

CCknP

nN

knMN

kM

MN ==−−

So, the general formula for calculation of the probability is:

( ) .k m km n m

mn

C CP А

C

−−

=

3.8 Suppose that there are 7 books. How many ways 4 books can be chosen

out of 7?

Solution:

47

7!.

4!(7 4)!С =

−

3.9 Suppose that 2 cards are drawn out from a well-shuffled deck of 52 cards.

What is the probability that both of them are spades?

Solution. The number of ways is n=52; of drawing out 2 cards from a well-

shuffled deck of 52 cards is 2

52С . Since 13 of the 52 cards are spades, the number of

ways m of drawing 2 spades is 2

13С . Thus:

P(getting 2 spades)=17

1

1326

782

52

2

13 ===C

C

n

m.

3.10 Suppose that 3 people are selected at random from a group that consists

of 6 men and 4 women. What is the probability that 1 man and 2 women are

selected?

Solution. The number of ways of selecting 3 people from a group of 10 is 310.C One man can be selected in

1

6C ways, and 2 women can be selected in 2

4C

ways. By the fundamental counting principle, the number of ways of selecting 1

27

man and 2 women is 2

4

1

6 CC . Thus the probability that 1 man and 2 women are

selected is:

10

33

10

2

4

1

6 =

=C

CCP .

4 Algebra of probabilities

Probability Properties. Probability P is the number, which satisfies three

conditions:

a) if an event A cannot occur, then P(A)=0;

b) if an event A is certain to occur, then P(A)=1;

c) the probability that an event A will occur is a number from 0 to 1:

0≤P(A)≤1.

Theorem (Probability of an Event). If A is an event in a sample space S and

nBBB ,...,, 21 are the individual outcomes comprising A, then:

=

=n

iiBPAP

1

)()( .

4.1 Assume all strings of three lowercase letters are equally probable. Then

what’s the probability of a string of three vowels?

Solution. There are 26 letters, of which 5 are vowels. So there are 326=n

three letter strings, and 35=m consisting only of vowels. Each outcome (string) is

equally likely, with probability n

1, so event A (a string of three vowels) has

probability:

.00711,026

5)(

3

3

==n

mAP

The Addition Law of Probability. As we have already noted the sample space

S is the set of all possible outcomes of a given experiment. Events A and B are

subsets of S. In the previous block we defined what was meant by P(A), P(B) and

their complements in the particular case in which the experiment had equally likely

outcomes.

Events, like sets, can be combined to produce new events:

- A∪B or A+B denotes the event that A or B (or both) occur when the

experiment is performed;

- A∩B or AB denotes the event that both A and B occur.

Two of the basic theorems of probability theory are connected with the

operations of union and intersection of events; these are the theorems of addition

and multiplication of probabilities.

28

Union / Addition. Let А and В bе two events. Then:

( ) ( ) ( ) ( ).P A B P A P B P A B = + +

The subtraction of )( BAP is necessary because А and В may «overlap». If

А аnd В аге mutually exclusive, i.e., = BA , then:

)()()( BPAPBAP += .

4.2 A bag contains 18 colored marbles: 4 are colored red, 8 are colored

yellow and 6 are colored green. A marble is selected at random. What is the

probability that the ball chosen is either red or green?

Solution. Assuming that any marble is as likely to be selected as any other we

can say:

- the probability that the chosen marble is red is 18

4 ;

- the probability that it is green is 18

6 .

It follows that the probability that the ball chosen is either red or green is

18

10

18

6

18

4=+ . This is the case when no ball can be simultaneously red and green. We

say that the events «the ball is red» and «the ball is green» are mutually exclusive.

Theorem (General Addition Rule). If A and B are two events in a sample

space S, then:

)()()()( BAPBPAPBAP −+= .

Of course if = BA then, since 0)( =P this general expression reduces to

the simpler version.

Ехаmрlе. Roll of а Dice:

- 6/3)( =evenP and ( 4) 3 / 6P less then = ;

- ( 4) ({2}) 1/ 6P even and less then P= = ;

- ( 4) 3 / 6 3 / 6 1/ 6 5 / 6P even or less then = + − = ;

- ({1} {6}) 1/ 6 1/ 6 0 2 / 6P or = + − = .

4.3 Consider a pack of 52 playing cards. A card is selected at random. What

is the probability that the card is either a diamond or a ten?

Solution. Let A be the event {a diamond is selected} and B be the event {a ten

is selected}. The probability that it is a diamond is 52

13)( =AP since there are 13

diamond cards in the pack. The probability that the card is a ten is 52

4)( =BP .

There are 16 cards that fall into the category of being either a diamond or a

ten: 13 of these are diamonds and there is a ten in each of the three other suits.

29

Therefore, the probability of the card being a diamond or a ten is 52

16 not 52

17

52

4

52

13=+ .

We say that these events are not mutually exclusive. We must ensure in this case

not to simply add the two original probabilities; this would count the ten of

diamonds twice - once in each category.

We used the addition law. This example 52

13)( =AP and

52

4)( =BP . The

intersection event A∩B consists of only one member - the ten of diamonds - hence

52

1)( = BAP . Therefore:

13 4 1 16( ) 0,3077

52 52 52 52P A B = + − = =

as we have already argued.

4.4 A bag contains 20 balls, 3 are colored red, 6 are colored green, 4 are

colored blue, 2 are colored white and 5 are colored yellow. One ball is selected at

random. Find the probabilities of the following events:

a) the ball is either red or green;

b) the ball is not blue;

c) the ball is either red or white or blue. (Hint: consider the complementary

event.)

Solution. Note that a ball can only have one colour, which are designated by

the letters R, G, B, W, Y:

a) 20

9

20

6

20

3)()()( =+=+= GPRPGRP ;

b) 5

4

20

41)(1)( =−=−= BPBP ;

c) the complementary event is G ∪ Y and:

20

11

20

5

20

6)()()( =+=+= YPGPYGP .

Hence:

11 9

( ) 1 0,45.20 20

P R W B = − = =

The answer in the last example (part (c)) we could alternatively have used an

obvious extension of the law of addition for mutually exclusive events:

3 2 4 9

( ) ( ) ( ) ( ) 0,45.20 20 20 20

P R W B P R P W P B = + + = + + = =

4.5 The following people are in a room: 5 men over 21, 4 men under 21, 6

women over 21, and 3 women under 21. One person is chosen at random. The

30

following events are defined: A={the person is over 21}; B={the person is under

21}; C={the person is male}; D={the person is female}. Evalute the following:

a) )( DBP ;

b) )( CAP . Express the meaning of these quantities in words.

Solution.

a) we have:

)()()()( DBPDPBPDBP −+= ;

18

7)( =BP ;

2

1

18

9)( ==DP ;

6

1

18

3)( == DBP ;

37 1 1 13

( ) 0,722318 2 6 18

P B D = + − = = .

b) find )( CAP . Let A ={people under 21}, C ={people who are female},

then:

6

1

18

3)( ==CAP .

4.6 A lot consists of 10 good articles, 4 with minor defects, and 2 with major

defects. One article is chosen at random. Find the probability that:

a) it has no defects (article is good);

b) it has no major defects;

c) it is either good or has major defects.

Solution. Let G={article is good}, Mj={major defect}, Mn={minor defect}.

Then:

a ) 10 5

( ) ;16 8

P G = = 10 5( ) ;

16 8P G = =

b) 5 4 7( ) ( ) ( ) ( ) 0 0,875;

8 16 8n n nP G M P G P M P G M = + − = + − = = ;

c) 5 2 6( ) ( ) ( ) ( ) 0 0,75.

8 16 8j j jP G M P G P M P G M = + − = + − = =

4.7 A card is drawn at random from a deck of 52 playing cards. What is the

probability that it is an ace or a face card?

Solution. Let F={face card}, A={card is ace}. Then:

;52

12)( =FP

52

4)( =AP .

We have:

31

12 4 16( ) ( ) ( ) ( ) 0 0,3077.

52 52 52P F A P F P A P F A = + − = + − = =

4.8 What is the probability that neither a double nor a 9 will appear in a single

throw of two dice?

Solution. Let D={double is thrown}, N={sum is 9}. Then there are 36

possible outcomes in an experiment in which all the outcomes are equally probable: 6

( ) ;36

P D =

1112,036

4)}63()54()45()36{()( === PNP ;

2778,036

100

36

4

36

6)()()()( ==−+=−+= NDPNPDPNDP ;

.7223,018

13

36

101)(1)( ==−=−= NDPNDP

4.9 (Independent events). A man owns a house in town and a cottage in the

country. In any one year the probability of the house being burgled is 0,01 and the

probability of the cottage being burgled is 0,05. What is the probability that in any

one year: both will be burgled? One or the other (but not both) will be burgled?

Solution. Let H={house is burgled}, C={cottage is burgled}:

P(H∩C)=P(H)P(C)=0,01 0,05=0,0005 since events independent.

P(one or the other (but not both)) = P((H∩C)∪(H∩C))=

= P(H∩C) + P(H∩C) = P(H)P(C) + P(H)P(C) =

= 0,01 0,95 + 0,99 0,05 = 0,059.

4.10 (Independent events). Current flows through a relay only if it is closed.

The probability of any relay being closed is 0.95. Calculate the probability that a

current will flow through a circuit composed of 3 relays in parallel. What

assumption must be made?

Solution. Let A be the event {relay A is closed}: Similarly for B, C req’d

event is:

1 23 3

{ } { } { }.

C C

A B C A B C A B C

Then:

32

P(Req`d event) 3 2 20,95 3 0,95 0,05 3 0,95 0,05 0,999875.= + + =

Conditional Probability. Definition (Conditional Probability, Joint

Probability). If A and B are two events in a sample space S, and 0)( AP then the

conditional probability of B given A is:

)(

)()/(

AP

BAPABP

= .

P(A∩B) is the joint probability of A and B, also written P(A,B). Intuitively,

P(B/A) is the probability that B will occur given that A has occurred.

Тhe reasoning behind this definition is that if В has осcurred, then only the

"portion" of А that is contained in В, i.e., BA , could occur; mоreovеr, the original

рrоbability of BA must bе recalculated to reflect the fact that the "new" sаmрlе

space is B:

Venn Diagram.

4.11 Pick а Саrd from а Deck. Suppose а card is drawn randomly from а deck

and found to bе an Асе. What is the conditional probability for this card to bе Spade

Асе?

Solution. А = Spade Асе; В = аn Асе; BA = Spade Асе:

52/1)(;52/4)(;52/1)( === BAPBPAP .

Неnсе:

.4

1

52/4

52/1)/( ==BAP

4.12 A manufacturer knows that the probability of an order being ready on

time is 0,80 and the probability of an order being ready on time and being delivered

on time is 0,72. What is the probability of an order being delivered on time, given

that it is ready on time?

Solution. Let R: order is ready on time; D: order is delivered on time.

P(R)=0,80; P(R,D)=0,72. Therefore:

33

.90,080,0

72,0

)(

),()/( ===

RP

DRPRDP

4.13 Consider sampling an adjacent pair of words (bigram) from a large text

T. Let

- BI = the set of bigrams in T (this is our sample space);

- A = «first word is run» = BITwwrun }:),{( 22 ;

- B=«second word is amok»= BITwamokw }:),{( 11 .

If 6,55,3 10)(,10)( −− == BPAP and 5,610),( −=BAP , what is the probability of

seeing amok following run, i.e., P(B/A)? How about run preceding amok, i.e.

P(A|B)?

Solution:

P(amok before run)=6,5

5,6

( , ) 10( / ) 0,126;

( ) 10

P A BP A B

P B

−

−= = =

P(amok after run)= .001,010

10

)(

),()/(

5,3

5,6

===−

−

AP

BAPABP

Independence. Two events аrе said to bе independent if the occurrence of

either оnе of the two events does not affect the occurrence probability of the other

event.

This is аn important concept, and is formаllу stated as: Two events А and В

are independent if:

)()/( APBAP = or ).()/( BPABP =

Note that the above is equivalent to:

)()()( BPAPBAP = .

Intuition: two events are independent if knowing whether one event occurred

does not change the probability of the other. Note that the following are equivalent:

P(A/B) = P(A);

P(B/A) = P(B).

4.14 A coin is flipped three times. Each of the eight outcomes is equally

likely. A: heads occurs on each of the first two flips, B: tails occurs on the third flip,

C: exactly two tails occur in the three flips. Show that A and B are independent, B

and C are dependent:

34

A = {HHH, HHT}, P(A) = 1/4;

B = {HHT, HTT, THT, TTT}, P(B) = 1/2;

C = {HTT, THT, TTH}, P(C) = 3/8;

A∩B = {HHT}, P(A∩B) = 1/8;

B∩C = {HTT, THT}, P(B∩C) = 1/4;

P(A)P(B) = 8

1

2

1

4

1= = P(A∩B).

Hence A and B are independent.

P(B)P(C) = =16

3

8

3

2

1 P(B∩C),

Hence B and C are dependent.

4.15 A simple example of two attributes that are independent: the suit and

value of cards in a standard deck: there are 4 suits {♦, ♠, ♣, ♥} and 13 values of

each suit {2,…,10, J, Q, K, A}, for a total of 52 cards.

Consider a randomly dealt card:

- marginal probability it’s a heart: P(suit=♥)=13/52 =1/4;

- conditional probability it’s a heart given that it’s a queen:

P(suit=♥/value=Q)=1/4;

- in general, P(suit/value) = P(suit), hence suit and value are independent.

4.16 We can verify independence by cross-multiplying marginal probabilities

too. For every suit:

s∈{♦, ♠, ♣, ♥}

and value:

v∈{2,…,10, J, Q, K, A}

we have:

- P(suit = s, value = v) = 1/52 (in a well-shuffled deck);

- P(suit = s) = 13/52 = 1/4;

- P(value = v) = 4/52 = 1/13;

- P(suit = s) · P(value = v) = 1 1 1.

4 13 52 =

4.17 Drawings with Replacement. Тwо balls аrе successively drawn from an

urn that contains sеvеn red balls and three black balls. The first bаll drawn is put

35

back into the urn after noting its color. What is the probability that the two balls

drawn аrе both black?

Solution. Let А - the first bаll drаwn is black; В - the sесоnd bаll drаwn is

black; BA - both balls аrе black; ;10/3)( =AP ,10/3)()/( == BPABP i.e. B and A are

independent. Неnсе:

.100

9

10

3

10

3)()/()( === APABPBAP

Independence comes up when we construct mathematical descriptions of our

beliefs about more than one attribute: to describe what we believe about

combinations of attributes, we often assume independence and simply multiply the

separate beliefs about individual attributes to specify the joint beliefs.

Conditional Independence.

Definition. Conditionally Independent Events. Two events A and B are

conditionally independent given event C iff:

P(A, B/C) = P(A/C)P(B/C).

Intuition: Once we know whether C occurred, knowing about A or B doesn’t

change the probability of the other.

The following are equivalent:

P(A, B/C) = P(A/C)P(B/C);

P(A/B, C) = P(A/C);

P(B/A, C) = P(B/C).

4.18 In a noisy room, I whisper the same number n∈{1,…,10} to two people

A and B on two separate occasions. A and B imperfectly (and independently) draw

a conclusion about what number I whispered. Let the numbers A and B think they

heard be an and bn , respectively.

Are an and bn independent (a.k.a. marginally independent)? No. E.g., we’d

expect P( an =1 / bn =1)>P( an =1).

Are an and bn conditionally independent given n? Yes: if you know the

number that I actually whispered, the two variables are no longer correlated. E.g., P(

an =1 / bn =1, n = 2) = P( an =1 / n = 2).

Multiplication. The multiplication rule is used to calculate the joint

probability of two events. It is simply а rearrangement of the conditional probability

formula:

36

( )( / )

( )

P A BP A B

P B

= .

Formally:

)()/()( BPBAPBAP =

or

)()/()( APABPBAP = .

Example. Drawing а Spade Асе. Let А - an Асе; В - а Spade; BA - the

Spade Асе:

13/1)/(;52/13)( == BAPBP .

Неnсе:

.52

1

52

13

13

1)()/()( === BPBAPBAP

4.19 Selecting students. А statistics course has seven male and three female

students. The professor wants to select two students at random to help her conduct а

research рrоject. What is the probability that the two students chosen аrе female?

Let А - the first student selected is female; В - the second student selected is

female; BA - both chosen students аге female:

.9/2)/(;10/3)( == ABPAP

Неnсе:

.15

1

10

3

9

2)()/()( === APABPBAP

Theorem of multiplication of probabilities. The probability of the intersection

of events rAAA ,...,, 21 is equal to the probability of event A1 multiplied by the

probability of event A2 under the condition that A1 has occurred, …, multiplied by

the probability of Ar under the condition that 121,...,,

−rAAA have occurred.

For independent events, the multiplication theorem reduces to the formula:

)(...)()()...(

2121 rrAPAPAPAAAP +++=

that is, the probability of the intersection of independent events is equal to the

product of the probabilities of these events.

Formula (3) remains correct, if on both sides some of the events are replaced

by their inverses.

37

4.20 Four shots are fired at a target, and the hit probability is 0,2 for each

shot. The target hits by different shots are assumed to be independent events. What

is the probability of hitting the target three times?

Solution. Each outcome of the trial can be designated by a sequence of four

letters [for example, (s, f, f, s) denotes that the first and fourth shots hit the target

(success), and the second and third miss (failure)].

There are 2 2 2 2 16 = outcomes in all. In accordance with the assumption

of independence of the results of individual shots, one should use formula (3) and

the remarks about it to determine the probabilities of these outcomes. Thus, the

probability of the outcome (s, f, f, f) is set equal to 1024,08,08,08,02,0 = ; here,

2,018,0 −= is the probability of a miss for a single shot. For the event «three shots hit

the target», the outcomes (s, s, s, f), (s, s, f, s), (s, f, s, s) and (f, s, s, s) are favorable

and the probability of each is the same:

0064,02,02,02,08,0...8,02,02,02,0 === .

Consequently, the desired probability is 4 0,0064 0,0256 = .

4.21 A bag contains 3 pink candies and 7 green candies. Two candies are

taken out from the bag with replacement. Find the probability that both candies are

pink.

Solution. Let A - event that first candy is pink and B - event that second candy

is pink:

P (A) =3/10.

Since the candies are taken out with replacement, this implies that the given

events A and B are independent:

P (B/A) = P (B) =3/10.

Hence by the multiplication law we get:

3 3

( ) ( ) ( / ) 0,009.10 10

P A B P A P B A = = =

4.22 A bag has 4 white cards and 5 blue cards. We draw two cards from the

bag one by one without replacement. Find the probability of getting both cards

white.

Solution. Let the total numbers of cards be 5 + 4 =9. Let A = event that first

card is white and B = event that second card is white.

38

From question, 4

( )9

P A = . Now )/()( ABPBP = , because the given events are

dependent on each other. 8

3)( =BP . So,

4 3 1( ) .

9 8 6P A B = =

Law of total probability. Тhe law of tota1 probability is the proposition that if

,...}3,2,1:{ =nBn is а finite or countably infinite partition of а sample space (in other

words, а set of pair wise disjoint events whose union is the entire sample space) and

each event }{ nB is measurable, then for any event A of the same probability space:

=n

nBAPAP )()(

or, altеnаtivеlу:

=n

nn BPBAPAP )()/()( .

Where for any n for which 0)( =nBP these terms are simply omitted from the

summation, because )/( nBAP is finite.

Тhe summation can bе interpreted as а weighted average, and consequently

the marginal probability, P(A) sometimes it is called «average рrоbаbility»;

sometimes «overall probability» is used in less formal writings.

The law of tota1 probability can also bе stated for conditional probabilities.

Taking the nB as above, and assuming C is an event independent with any of the

nB :

( / ) ( / ) ( / )n n

n

P A C P A C B P B C= .

4.23 Suppose that two factories supply light bulbs to the market. Factory X's

bulbs work for over 5000 hours in 99% of cases, whereas factory Y's bu1bs work for

over 5000 hours in 95% of cases. It is known that factory Х supplies 60% of the

tota1 bu1bs available. What is the chance that а purchased bulb will work for longer

than 5000 hours?

Solution. Applying the law of tota1 probability, we have:

( ) ( / ) ( ) ( / ) ( )X X Y YP A P A B P B P A B P B= + =

99 6 95 4 594 380

0,974,100 10 100 10 1000

+= + = =

where

39

- 10

6)( =XBP is the probability that the purchased bulb was manufactured bу

factory Х;

- 10

4)( =YBP is the probability that the purchased bulb was manufactured bу

factory Y;

- 100

99)/( =XBAP is the probability that а bu1b manufactured bу Х will work

for over 5000 hours;

- 100

95)/( =YBAP is the probability that а bu1b manufactured bу Y will work

for over 5 000 hours.

Thus each purchased light bu1b has а 97,4% chance to work for more than

5000 hours.

The term law of total probability is sometimes taken to mean the law of

alternatives, which is а special case of the law of tota1 probability applying to

discrete random variables.

4.24 А disease called pluremia affects 1% of the population. There is а test to

detect pluremia but it is not perfect. For people with pluremia, the test is positive

90% of the time. For people without pluremia the test is positive 20% of the time.

Suppose а randomly selected person takes the test and it is positive. What are the

chances that а randomly selected person tests positive?

Solution. Let A represent а positive test result, B not having pluremia, C

having pluremia. We know Р(B)=0,99; Р(C)=0,01. Тhе test specifications tell us:

P(A/B)=0,2 and P(A/C)=0,9. Applying the LTP:

( ) ( ) ( )P A P A B P A C= + =

.207,09,001,02,099,0)/()()/()( =+=+= CAPCPBAPBP

Joint Probability and Multiplication Rule. From the definition of conditional

probability, we obtain:

Theorem (Multiplication Rule). If A and B are two events in a sample space S

and 0)( AP , then:

)./()(),( ABPAPBAP =

Since A∩B=B∩A, we also have that:

)./()(),( BAPBPBAP =

Marginal Probability and the Rule of Total Probability.

40

Theorem: Marginalization (a.k.a. Rule of Total Probability). If events

kBBB ,...,, 21 constitute a partition of the sample space S and 0)( iBP for i=1,2,...,k,

then for any event A in S:

= =

==k

i

k

iiii BPBAPBAPAP

1 1

)()/(),()( .

Marginalization. kBBB ,...,, 21 form a partition of S if they are pairwise

mutually exclusive and if:

SBBB k = ...21 .

4.25 In an experiment on human memory, participants have to memorize a set

of words ( 1B ), numbers )( 2B and pictures ( 3B ). These occur in the experiment with

the probabilities:

.1,0)(;4,0)(;5,0)( 321 === BPBPBP

Then participants have to recall the items (where A is the recall event). The

results show that:

.1,0)/(;2,0)/(;4,0)/( 321 === BAPBAPBAP

Compute P(A), the probability of recalling an item.

By the theorem of total probability:

1

( ) ( ) ( / )k

i i

i

P A P B P A B=

= =

1 1 2 2 3 3( ) ( / ) ( ) ( / ) ( ) ( / )P B P A B P B P A B P B P A B= + + =

0,5 0,4 0,4 0,2 0,1 0,1 0,29.= + + =

4.26 (Independent events). A box contains 4 bad and 6 good tubes. Two are

drawn out together. One of them is tested and found to be good. What is the

probability that the other one is also good?

Solution. Let Gi = {i-th tube is good}, Bi = {i-th tube is bad}. Then

9

5)/( 12 =GGP (only 5 good tubes left out of 9).

4.27 (Independent events). In the above problem the tubes are checked by

drawing a tube at random, testing it and repeating the process until all 4 bad tubes

41

are located. What is the probability that the fourth bad tube will be located: (a) on

the fifth test? (b) on the tenth test?

Solution. Same events as in the last example:

(a) This will occur if event:

{B1∩B2∩B3∩G4∩B5}∪{...∩B5}∪{...B5}∪{...B5}

occurs. Here we have a number of events in which B5 must appear in the last

position and there must be just three appearances of the B symbol in the first 4 slots.

Now the number of ways of arranging 3 from 4 is 34Ñ .

Thus the probability of the required event occurring is 34Ñ

P{B1∩B2∩B3∩G4∩B5}. Then:

P(event occuring) = 34Ñ P{B1∩B2∩B3∩G4∩B5}=

= 34Ñ P(B5/(B1∩B2∩B3∩G4)) P(B1∩B2∩B3∩G4) =

= 34Ñ P(B5/(B1∩B2∩B3∩G4)) P(G4/(B1∩B2∩B3)) …

… P(B3/(B1∩B2)) P(B2/B1) P(B1) =

=105

2

210

4

10

4

9

3

8

2

7

6

6

134 ==Ñ .

(b) Same idea as in (a) :

Req’d probability =

= 39Ñ P(B10/B1∩B2∩...∩G9)P(B1∩...∩G9) =

= .4,05

2

10

4

9

3

8

2

7

6

6

5

5

4

4

3

3

2

2

1

1

139 ==Ñ

4.28 (Independent events). Current flows through a relay only if it is closed.

The probability of any relay being closed is 0.95. Calculate the probability that a

current will flow through a circuit composed of 3 relays in parallel. What

assumption must be made?

Solution. Let A be event {relay A is closed}: Similarly for B, C req’d event is:

1 23 3

{ } { } { }.

C C

A B C A B C A B C

Then:

42

P(Req`d event) =

.999875,005,095,0305,095,0395,0 223 =++=

Joint, Marginal Probability and Conditional Probability

4.29-1 Proportions for a sample of University students, N=592.

Hair Color

Eye Color black brunette blond red

blue 0,03 0,14 0,16 0,03 0,36

brown 0,12 0,20 0,01 0,04 0,37

hazel/green 0,03 0,14 0,04 0,05 0,27

0,18 0,48 0,21 0,12

4.29-2 These are the joint probabilities:

P(eye Color, hair Color).

Hair Color

Eye Color black b

runette

b

lond

r

ed

blue 0,03 0,14 0,16 0,03 0,36

brown 0,12 0,20 0,01 0,04 0,37

hazel/green 0,03 0,14 0,04 0,05 0,27

0,18 0,48 0,21 0,12

4.29-3 E.g.:

P(eye Color = brown, hair Color = brunette) = 0,20.

Hair Color

Eye Color black brunette blond red

blue 0,03 0,14 0,16 0,03 0,36

brown 0,12 0,20 0,01 0,04 0,37

hazel/green 0,03 0,14 0,04 0,05 0,27

0,18 0,48 0,21 0,12

4.29-4 These are the marginal probabilities:

P(eye Color).

Hair Color

Eye Color black brunette blond red

43

blue 0,03 0,14 0,16 0,03 0,36

brown 0,12 0,20 0,01 0,04 0,37

hazel/green 0,03 0,14 0,04 0,05 0

,27

0,18 0,48 0,21 0,12

4.29-5 E.g.:

P(eye Color = brown) =

( , )hairColor

P eyeColor brown hairColor= = =

0,12 0,20 0,01 0,04 0,37.= + + + =

Hair Color

Eye Color black brunette blond red

blue 0,03 0,14 0,16 0,03 0,36

brown 0,12 0,20 0,01 0,04 0,37

hazel/green 0,03 0,14 0,04 0,05 0,27

0,18 0,48 0,21 0,12

4.29-6 These are the marginal probabilities P(hair Color).

Hair Color

Eye Color black brunette blond red

blue 0,03 0,14 0,16 0,03 0,36

brown 0,12 0,20 0,01 0,04 0,37

hazel/green 0,03 0,14 0,04 0,05 0,27

0,18 0,48 0,21 0,12

4.29-7 E.g.:

P(hairColor = brunette) =

( , )eyeColor

P eyeColor hairColor brunette= = =

0,14 0,20 0,14 0,48.= + + =

Hair Color

Eye Color black brunette blond red

44

blue 0,03 0,14 0,16 0,03 0,36

brown 0,12 0,20 0,01 0,04 0,37

hazel/green 0,03 0,14 0,04 0,05 0,27

0,18 0,48 0,21 0,12

4.29-8 To obtain the conditional probability:

P(eye Color / hair Color = brunette),

we do two things:

Hair Color

Eye Color black brunette blond red

blue 0,03 0,14 0,16 0,03 0,36

brown 0,12 0,20 0,01 0,04 0,37

hazel/green 0,03 0,14 0,04 0,05 0,27

0,18 0,48 0,21 0,12

4.29-9 To obtain the conditional probability:

P(eye Color / hair Color = brunette),

we do two things:

1) Reduction: we consider only the probabilities in the brunette column:

Hair Color

Eye Color black brunette blond red

blue 0,14

brown 0,20

hazel/green 0,14

0,48

To obtain the conditional probability

P(eye Color / hair Color = brunette),

we do two things:

2) Normalization: we divide by the marginal P(brunette), since all the

probability mass is concentrated here now:

Hair Color

Eye Color black brunette blond red

blue 0,14/0,48

brown 0,20/0,48

hazel/green 0,14/0.48

0,48

4.29-10 E.g.:

45

P(eye Color = brown / hair Color = brunette) =0,20/0,48.

Hair Color

Eye Color black brunette blond red

blue 0,14/0,48

brown 0,20/0,48

hazel/green 0,14/0.48

0,48

4.29-11 Moreover:

P(eye Color = brown / hair Color = brunette)

P(hair Color = brunette / eye Color = brown).

Consider P(hair Color / eye Color = brown):

Hair Color

Eye Color black brunette blond red

blue 0,03 0,14 0,16 0,03 0,36

brown 0,12 0,20 0,01 0,04 0,37

hazel/green 0,03 0,14 0,04 0,05 0,27

0,18 0,48 0,21 0,12

4.29-12 To obtain P(hair Color / eye Color = brown), we reduce:

Hair Color

Eye Color black brunette blond red

blue

brown 0,12 0,20 0,01 0,04 0,37

hazel/green

and we normalize:

Hair Color

Eye Color black brunette blond red

blue

brown 0,12/0,37 0,20/0,37 0,01/0,37 0,04/0,37 0,37

hazel/green

46

So:

P(hair Color = brunette / eye Color = brown) = 0,20/0,37,

Hair Color

Eye Color black brunette blond red

blue

brow 0,12/0,37 0,20/0,3 0,01/0,37 0,04/0,37 0,37

hazel/green

But:

P(eye Color = brown / hair Color = brunette) = 0,20/0,48.

Hair Color

Eye Color black brunette blond red

blue 0,14

brown 0,20/0,48

hazel/green 0,14

0,48

4.30 There are two urns with balls. In the first -2 white and 4black, the

second-3 white and 2 black. From each urn at random heed one ball. Find the

probability that: a) the balls have the same color; b) at least one of them is white

ball.

4.31 In the first box there are 32% of the parts of the first grade, in the second

- 25%. They each box randomly heed one piece. Find the probability that among

them: a) there is only one part of the first grade b) at least there is one piece of first

grade

4.32 In the first urn there are 2 white and 10 black balls; the second - 8 white

and 4 black balls. From each ballot box it was taken out of the ball. Find the

probability that among them: a) only one is black; b) at least one is black.

4.33 Three shooters produce one shot on goal, hit probability of which are

equal to 0,5 for the first; 1,7 for the second; 0,8 - for the third. Find the probability

that: a) two hit the target; b) at least one hits the target.

4.34 The probability of increasing the price of milk in the current month is

0,8; for meat – 0,65. Find the probability that in the current month price will rise: a)

only for one product; b) at least for one product.

4.35 A car is equipped with two anti-theft devices, the likelihood of

disconnection of which is 0,04 and 0,06 respectively. Find the probability of

inclusion in the hijacking: a) only one device; b) at least one.

47

4.36 Two shooters produce one shot on target, the likelihood of which is

equal to 0,85 for the first shooter; 0,6 - for the second. Find the probability that: a)

only one hits the target; b) at least one.

4.37 There are two boxes, each containing 10 parts. There are two non-

standard parts in the first; three in the second. One piece is randomly removed from

each box. Find the probability that standard would be: a) only one item; b) at least

one item.

4.38 In the first box 55% of the parts are of the first grade, the second - 40%.

One piece is randomly removed from each box. Find the probability that there is: a)

only one piece of the first grade; b) at least one piece of the second grade.

4.39 The device consists of two independently operating unit, which

probabilites of failure as a result of the test are, respectively, 0.25 and 0.3. Find the

probability of failure as a result of the test: a) only of one block; b) at least of one

block.

4.40 At the exam the student received a ticket containing 3 questions.

Probabilities of correct answers to the questions of the ticket are respectively 0.9,

0.8, 0.7. Find the probability that a student answers correctly: a) 2 questions; b) at

least one question.

4.41 The shop is equipped with two alarm that triggered an accident with

probabilities 0.9 and 0.85. Find the probability that an accident will work: a) only

one alarm; b) at least one indicator.

4.42 Three guns shoot at the target. The probabilities of hitting the target in

one shot for these guns are respectively 0.7, 0.8 and 0.9. Find the probability that

the result will be: a) only one; b) at least one; c) three

4.43 Two boxes have the details: in the first - 10, including 8 standard; the

second - 15, 9 of which are standard. One piece randomly removed from each box.

Find the probability that the standard will be: a) two parts; b) only one detail; c) at

least one item.

4.44 Two shooters shot at the target. The probability of hitting the target is –

0,7 for the first, for the second – 0,65. Find the probability that the target will be

destroyed: a) only by one shooter; b) at least by one shooter.

Bayes’ Theorem. In probability theory Bayes' theorem (Bayes' law or

Bayes' rule) describes the probability of an event, based on conditions that might be

related to the event.

If nAAA ,...,, 21

are mutually exclusive events:

SAAA n = ...21

and B is another event, then Bayes' theorem is stated mathematically as the

following equation:

48

1 1 2 2

( ) ( / )( / )

( ) ( / ) ( ) ( / ) ... ( ) ( / )i i

i

n n

P A P B AP A B

P A P B A P A P B A P A P B A

=

+ + +

.

Where:

- )( iAP - is the prior probability;

- )/( BAP i - is the posterior probability.

4.45 While watching a game of Champions League football in a cafe, you

observe someone who is clearly supporting Manchester United in the game. What is

the probability that they were actually born within 25 miles of Manchester?

Solution. Assume that:

- the probability that a randomly selected person in a typical local bar

environment born within 25 miles of Manchester is 1/20, and 19/20;

- the chance that a person born within 25 miles of Manchester actually

supports United is 7/10;

- the probability that a person not born within 25 miles of Manchester

supports United with probability 1/10 .

Define

- B - event that the person is born within 25 miles of Manchester;

- U - event that the person supports United.

We want P(B/U). By Bayes’ Theorem:

=+

=

=

)()/()()/(

)()/(

)(

)()/()/(

BPBUPBPBUP

BPBUP

UP

BPBUPUBP

.269,020

7

20

19

10

1

20

1

10

720

1

10

7

=

+

=

4.46 A fair coin is flipped three times. There are 8 possible outcomes, and

each of them is equally likely.

For each outcome, we can count the number of heads and the number of

switches (i.e., HT or TH subsequences):

outcome probability #heads #switches

HHH 1/8 3 0

THH 1/8 2 1

HTH 1/8 2 2

HHT 1/8 2 1

TTH 1/8 1 1

THT 1/8 1 2

HTT 1/8 1 1

TTT 1/8 0 0

49

4.47-1 The joint probability P(#heads, #switches) is therefore:

#heads

#switches 0 1 2 3

0 1/8 0 0 1/8 2/8

1 0 2/8 2/8 0 4/8

2 0 1/8 1/8 0 2/8

1/8 3/8 3/8 1/8

Let us use Bayes’ theorem to relate the two conditional probabilities:

P(#switches = 1 / #heads = 1);

P(#heads = 1 / #switches = 1).

4.47-2

#heads

#switches 0 1 2 3

0 1/8 0 0 1/8 2/8

1 0 2/8 2/8 0 4/8

2 0 1/8 1/8 0 2/8

1/8 3/8 3/8 1/8

Note that:

P(#switches = 1 / #heads = 1) = 2/3;

P(#heads = 1 / #switches = 1) = 1/2.

4.47-3

#heads

#switches 0 1 2 3

0 1/8 0 0 1/8 2/8

1 0 2/8 2/8 0 4/8

2 0 1/8 1/8 0 2/8

1/8 3/8 3/8 1/8

The joint probability:

P(#switches = 1, #heads = 1) = 2/8

can be expressed in two ways:

50

P(#switches = 1 / #heads = 1) P(#heads = 1) = .8

2

8

3

3

2=

4.47-4

#heads

#switches 0 1 2 3

0 1/8 0 0 1/8 2/8

1 0 2/8 2/8 0 4/8

2 0 1/8 1/8 0 2/8

1/8 3/8 3/8 1/8

The joint probability:

P(#switches = 1, #heads = 1) = 2/8

can be expressed in two ways:

P(#heads = 1 / #switches = 1)·P(#switches = 1) = .8

2

8

4

2

1=

4.47-5

#heads

#switches 0 1 2 3

0 1/8 0 0 1/8 2/8

1 0 2/8 2/8 0 4/8

2 0 1/8 1/8 0 2/8

1/8 3/8 3/8 1/8

Bayes’ theorem is a consequence of the fact that we can reach the joint:

P(#switches = 1, #heads = 1).

In these two ways:

- by restricting attention to the row #switches = 1;

- by restricting attention to the column #heads = 1.

4.48 20% of a company's employees are engineers and 20% are economists.

75% of the engineers and 50% of the economists hold a managerial position, while

only 20% of non-engineers and non-economists have a similiar position. What is the

probability that an employee selected at random will be both an engineer and a

manager?

0,2 0,75

( / ) 0,405.0,2 0,75 0,2 0,5 0,6 0,2

P engineer managerial

= = + +

51

4.49 The probability of having an accident in a factory that triggers an alarm

is 0,1. The probability of it sounding after the event of an incident is 0,97 and the

probability of it sounding after no incident has occured is 0,02. In an event where

the alarm has been triggered, what is the probability that there has been no accident?

Solution. I = Accident occurred:

0,1; 1 0,1 0,9I I= = − = .

A = Triggered alarm:

0,97; 1 0,97 0,03A A= = − = .

0,9 0,02

( / ) 0,157.0,1 0,97 0,9 0,02

P I A

= = +

4.50 For a hypothetical magazine, the probability that the reader is male,

given that the reader is at least 35 years old, is 0,30. The probability that a reader is

male, given that the reader is under 35, is 0,65. If 75% of the reader are under thirty

five what is the probability that a randomly chosen reader is:

a) male;

b) female;

c) under 35;

if it is given the reader is a female.

Solution:

a) let A1 be the event of the reader being at least 35 years and A2 the event of

the reader being under 35 years old and B the event of the reader being a male:

P(A2) = 0,75 and P(A1) = 1 - P(A2) = 1 - 0,75 = 0,25;

P(B/A1) = 0,30 and P(B/A2) = 0,65.

52

B = (A1∩B)U(A2∩B) as the male reader can either under 35 or at least 35

years old:

P(B) = P(A1∩B) + P(A2∩B) = P(A1);

P(B/A2) = (0,25)(0,30) + (0,75)(0,65) = 0,5625.

The probability the reader is male =0,5625;

b) P(female) = 1 - P(male) = 1 - 0,5625 = 0,4375. The probability the reader

is female = 0,4375;

c) let E1 be the event of the reader is under 35 and E2 the event of the reader is

at least 35 and the F the event of the reader being a female.

We need to find P(E1/F) and we have the probabilities:

P(F/E1) = 1 - 0,65 = 0,35; P(F/E2) = 1 - 0,30 = 0,70;

P(E1) = 0,75 and P(E2) = 0,25;

1 11

1 1 2 2

( / ) ( )( / )

( / ) ( ) ( / ) ( )

P F E P EP E F

P F E P E P F E P E= =

+

0,75 0,350,6.

0,75 0,35 0,25 0,70

= =

+

Probability the reader is under 35 given that the reader is a female = 0,6.

4.51 The probability the event A will occur is 0,4. The conditional probability

of another event B occurring given A is 0,6 and the conditional probability of B

occurring given A has not occurred is 0,3. Find the probability of:

a) the event A occurring given B;

b) the event of A not occurring given B.

Solution. Expressing the given probabilities using notation 4,0)( =AP . Hence:

6,04,01)( =−=AP ; 4,0)/( =ABP and 3,0)/( =ABP .

a) We need to compute P (A/B). Using Bayes' Theorem:

.571,03,06,06,04,0

6,04,0

)()/()()/(

)()/()/(

=+

=

=+

=APABPAPABP

APABPBAP

Probability of event A occurring given B = 0,571.

b) Let`s find:

53

429,0571,01)/(1)/( =−=−= BAPBAP .

Probability of event A not occurring given B = 0,429.

4.52 In a TV Game show, a contestant selects one of three doors; behind one

of the doors there is a prize, and behind the other two there are no prizes. After the

contestant selects a door, the game-show host opens one of the remaining doors, and

reveals that there is no prize behind it. The host then asks the contestant whether

they want to SWITCH their choice to the other unopened door, or STICK to their

original choice.

Is it probabilistically advantageous for the contestant to SWITCH doors, or is

the probability of winning the prize the same whether they STICK or SWITCH?

(Assume that the host selects a door to open, from those available, with equal

probability).

Solution. Without loss of generality, let events A, B, C correspond to the prize

being behind the selected, opened, and remaining door respectively, and let HB

denote the event that the host opens door B. Want to compare P(A/HB) (STICK)

with P(C/HB) (SWITCH). Now:

P(A) = P(B) = P(C) = 1/3,

and we are given that:

P(HB/A) = 1/2, P(HB/B) = 0 and P(HB/C) = 1.

Then the general version of Bayes’ theorem gives:

( / ) ( )

( / )( )

BB

B

P H A P AP A H

P H= =

( / ) ( )

( / ) ( ) ( / ) ( ) ( / ) ( )B

B B B

P H A P A

P H A P A P H B P B P H C P C= =

+ +

1 112 3 .

1 1 1 1 30 1

2 3 3 3

= =

+ +

So:

P(A/HB) = 1/3,

54

and similarly:

P(C/HB) = 2/3.

So it is advantageous to SWITCH.

4.53 A diagnostic test has a probability 0.95 of giving a positive result when

applied to a person suffering from a certain disease, and a probability 0,10 of giving

a (false) positive when applied to a non-sufferer. It is estimated that 0,5% of the

population are sufferers. Suppose that the test is now administered to a person about

whom we have no relevant information relating to the disease (apart from the fact

that he/she comes from this population). Calculate the following probabilities:

a) that the test result will be positive;

b) that, given a positive result, the person is a sufferer;

c) that, given a negative result, the person is a non-sufferer;

(d) that the person will be misclassified.

Solution. Let:

T ≡ «Test positive», S ≡ «Sufferer», M ≡ «Misclassified».

Then:

P(T/S) = 0,95; P(T/S) = 0,10; P(S) = 0,005.

Hence:

) ( ) ( / ) ( ) ( / ) ( ) 0,95 0,005 0,1 0,995 0,10425;a P T P T S P S P T S P S= + = + =

( / ) ( )

) ( / )( / ) ( ) ( / ) ( )

P T S P Sb P S T

P T S P S P T S P S= =

+

0,95 0,005

0,0455;0,95 0,005 0,1 0,995

= =

+

( / ) ( ) 0,9 0,995) ( / ) 0,9997;

1 0,10425( )

P T S P Sc P S T

P T

= = =

−

) ( ) ( ) ( )

( / ) ( ) ( / ) ( ) 0,09975.

d P M P T S P T S

P T S P S P T S P S

= + =

= + =

4.54 А disease called pluremia affects 1% of the population. There is а test to

detect pluremia but it is not perfect. For people with pluremia, the test is positive

90% of the time. For people without pluremia the test is positive 20% of the time.

Suppose а randomly selected person takes the test and it is positive. What are the

chances that this person has pluremia?

Let A represents а positive test result, B not having pluremia, C having

pluremia. Prior to the probabilities of B and C are as follows: Р(B) = 0,01, P(C) =

55

0,99.

The test specifications give us the following probabilities: P(A/C) = 0,9 and

P(A/B) = 0,2.

Applying Bayes' theorem:

0,9 0,01

( / ) 0,043.0,9 0,01 0,2 0,99

P C A

= = +

Кnowing that the test is positive increases the chances of having pluremia

from 1 in а hundred to 4,3 in а hundred.

4.55 А doctor believes that а patient has а 10% сhаnсе of having Lyme

disease. She gives the patient а blood test and the test comes out positive. Тhе

mаnuа1 for this test says that that out of 100 patients with Lyme disease, 80% of

tests are positive. Moreover, out of 100 patients with по Lyme disease 30% of tests

are positive. What is the probability that the patient has Lyme disease?

Solution. If you were а pure classic you would bе unwilling to answer this

question. Yоu would simply say that probabilities should not bе applied to empirical

hypotheses. This person either has Lyme disease or hе does not. You would simply

say that your tools do not apply to this problem.

If you were Bayesian you would bе willing to use probability theory to play

with knowledge and internal beliefs so this question makes sense to you. You could

represent the hypothesis that the patient has Lyme disease as аn event A which has а

prior probability of 10%:

9,0)(1)(;1,0)( =−== BPCPBP .

Where C represents the hypothesis that the patient does not have Lyme

disease.

The positive test is а data event A with the following characteristics:

3,0)/(;8,0)/( == CAPBAP .

Now уоu could apply Bayes' theorem to compute the probability of the

hypotheses given the data

.23,09,03,01,08,0

1,08,0)/( =

+

=ABP

After seeing the results of the test, it would bе rational for the Doctor to

update her beliefs and give her patient а 23% probability of having Lyme disease.

Note the emphasis here is оn upgrading beliefs based оn empirical data. The

emphasis is not оn deciding whether а hypothesis is true or fa1se. Of course it is

now up to the doctor and her patient to use the 23% probability figure to perhaps get

а better test or to evaluate the costs and benefits of treatments.

56

4.56 The entire output of а factory is produced оn three machines. Three

machines account for 20%, 30%, and 50% of the output, respectively. The fraction

of defective items produced is this: for the first machine - 5%; for the second

machine - 3%; for the third machine - 1%. If an item is chosen at random from the

total output and is found to bе defective, what is the probability that it was produced

bу the third machine?

Solution. Let iA denote the event that а randomly chosen item was made bу

the i-th machine (for i=1,2,3). Let В denote the event that а randomly chosen item

is defective. Then, we are given the following information:

1 2 3( ) 0,2; ( ) 0,3; ( ) 0,5.P A P A P A= = =

If the item was made bу machine 1A , then the probability that it is defective

is 0,05; that is, .05,0)/( 1 =ABP Overall, we have:

.01,0)/(;03,0)/(;05,0)/( 321 === ABPABPABP

То answer the original question, we first find Р(В). That can bе done in the

following way

( ) ( / ) ( )i i

i

P B P B A P A= =

0,05 0,2 0,03 0,3 0,01 0,5 0,024.= + + =

Hence 2,4% of the total output of the factory is defective.

Wе are given that В has occurred, and we want to calculate the conditional

probability of 3A . Ву Bayes' theorem:

.24

5

024,0

50,001,0

)(

)()/()/( 33

3 =

=

=BP

APABPBAP

Given that the item is defective, the probability that it was made bу the third

machine is only 5/24. Although machine 3 produces half of the total output, it

produces а much smaller fraction of the defective items.

Hence the knowledge that the item selected was defective enables us to

replасе the prior probability 2/1)( 3 =AP bу the smaller posterior probability

.24/5)/( 3 =BAP

Once again, the answer can bе reached without recourse to the formula bу

applying the conditions to any hypothetical number of cases. For example, in

100,000 items produced bу the factory: 20,000 will bе produced bу Machine А;

30,000 bу Machine В; and 50,000 bу Machine С. Machine А will produce 1000

defective items, Machine В - 900 and Machine С - 500. Of the total 2400 defective

57

items, only 500, or 5/24 were produced bу Machine С.

4.57 Briefly describe in your own words the difference between the

frequency, Bayesian and mathematical notions of probability.

4.58 Urn А has 3 black balls and 6 white balls. Urn В has 400 black balls and

400 white balls. Urn С has 6 black balls and 3 white balls. А person first randomly

chooses one of the urns and then grabs а bаll randomly from the chosen urn. What

is the probability that the bаll bе black? If а person grabbed а black bаll. What is the

probability that the bаll camе from urn В?

4.59 The probability of catching Lyme disease after оn day of hiking in the

Сuуаmаса mountains are estimated at less than 1 in 10000. Yоu feel bad after а day

of hike in the Cuyamaca and decide to take а Lyme disease test. The test is positive.

The test specifications say that in an experiment with 1000 patients with Lyme

disease, 990 tested positive. Moreover when the same test was performed with 1000

patients without Lyme disease, 200 tested positive. What are the chances that you

got Lyme disease.

4.60 This рroblem uses Bayes' theorem to combine probabilities as subjective

beliefs with probabilities as relative frequencies. А friend of yours believes she has

а 50% chance of being pregnant. She decides to take а pregnancy test and the test is

positive. Yоu read in the test instructions that out of 100 non-pregnant women, 20%

give false positives. Moreover, out of 100 pregnant women 10% give false

negatives. Help your friend upgrade her beliefs.

4.61 Тhere are three boxes, each containing а different number of light bulbs.

The first bох has 10 bu1bs, of which four arе dead, the second has 6 bu1bs, of

which one is dead, and the third bох has 8 bu1bs of which three arе dead. What is

the probability of а dead bu1b being selected when а bulb is chosen at random from

one of the three boxes?

4.62 There are three urns containing Black, Red and Green Balls. Urn 1

contains 4 Black, 5 Red and 3 Green Balls. Urn 2 contains 7 Black, 4 Red and 4

Green balls. Urn 3 contains 6 Black, 6 Red and 5 Green Balls. A Urn is chosen and

a ball is picked. If the ball picked is Green find the probability it has come from Urn

3. (Answer: 0,363).

4.63 The probability of the incidence of a rare disease is 0,002. The

probabilities of diagnostic disease showing positive are 0,97 and 0,05 for an

infected and not an infected situations. When the test turns out positive, what is the

probability the person tested is really has the disease. (Answer: 0,0374).

4.64 The collector received 3 boxes of parts manufactured at plant No. 1, and

2 boxes of parts manufactured at plant No. 2. The probability that the item of plant

No. 1 standard, equal to 0.8, and the part of plant No. 2 is 0.9. The collector

randomly pulled the item from the box at random taken. Find the probability that

the extracted standard detail.

4.65 At an observation station installed radar 3 different designs. The

probability of target detection by using the first locator is equal to 0,86, second –

0,92, third – 0,95. The observer turns off one of the locators. Find the probability of

target detection.

58

4.66 There are five rifles, three of which with an optical sight. The probability

of hitting with one shot from a rifle with a telescopic sight for a given hand is 0.95,

with a rifle without scope for 0.7. Find the probability of hitting the target are taken

at random from the rifle.

4.67 The company receives components from two suppliers. Delivery

volumes are as 2:1. In the production of the first provider 96% of standard products,

the second – 93%. Selective control has become faulty. Find the probability that the

marriage happened because of the second supplier.

4.68 Products are to check one of the Ghost controllers. The likelihood that

products to the first controller are equal to 0,6 and the second is 0,4. The likelihood

that a suitable product is found standard the first control equal to 0,95 and the

second – 0,9. Find the probability that the product test will be recognized as

standard.

4.69 Among 350 mechanisms there are 160 of the first class, 110 second

class, 80 third grade. The probability of marriage arrangements for the first grade is

0.01, for mechanisms of the second grade is 0.02, mechanisms for third grade is

0.04. Taken one mechanism. Find the probability that this mechanism is in order.

4.70 9 white and 5 black balls in the first urn, in the second - three white and

seven black balls. From the second ballot box in the first shifted one ball and then

the urn was taken out from the first one ball. Find the probability that a white ball

was taken.

4.71 The pilot catapulted in the area, 60% of which is covered by forests. The

probability of successful pilot landing in the forest is 0,3 and in treeless areas – 0,9.

Find the probability of successful pilot landing.

4.72 Clothes arrive in stores in the city with three factories, 40% from the

first, 30% - in the second, 30% - with the third. The number of products having

latent defects is 1% for the first factory, 3% - for the second, 5% - for the third. Find

the probability that the store bought no defects thing.

4.73 In the shop there are 3 and 6 masters of their students. Master admits

marriage with a probability of 0,06; the student - with a probability of 0,21. It came

from the shop the product is defective. Find the probability that it produced the

master.

4.74 In the shops of the city received the same type of goods from three

plants, with the first plant supplies 50% of the products, the second - 30%, the third

- 20%. Among the first products of the plant 70% of first-class, second - 80%, the

third - 90%. Bought one product. It turned out to be top-notch. Find the probability

that it released the first plant.

4.75 Each of the two boxes contain 3 black and 7 white balls. From the first

ballot boxes randomly removed one ball and shift the second, and then removed

from the second ball of the urn. Find the probability that the ball will be white.

4.76 The vase light bulbs are manufactured in two plants. Of these, 60% is

made on the first plant, 40% - in the second. The probability that the lamp complies

with the standard, is equal to 0.9 for the first plant, 0.8 - second plant. It takes the

59

guesswork proved a standard bulb. Find the probability that it is made in the first

plant.

4.77 In box 10 revolvers single systems and identical in appearance: 4 of

them did not adjust. The probability of not getting sighted revolver is 0.3, and from

sighted - 0.9. Find the probability of hitting the target of random selected revolver.

4.78 In the area of products supplied by two companies in the ratio of 2: 3.

Among the first products of the company standard products account for 90%, the

second - 85%. Take the guesswork proved to the standard product. Find the

probability that it is made by the first.

4.79 At an observation station installed radar 3 different designs. The

probability of target detection by using the first locator is equal to 0,86, second –

0,92, third – 0,95. The observer turns off one of the locators. Find the probability of

target detection.

4.80 There are five rifles, three of which with an optical sight. The probability

of hitting with one shot from a rifle with a telescopic sight for a given hand is 0.95,

with a rifle without scope for 0.7. Find the probability of hitting the target are taken

at random from the rifle.

4.81 The company receives components from two suppliers. Delivery

volumes are as 2:1. In the production of the first provider 96% of standard products,

the second – 93%. Selective control has become faulty. Find the probability that the

marriage happened because of the second supplier.

4.82 Products are to check one of the Ghost controllers. The likelihood that

products to the first controller are equal to 0,6 and the second is 0,4. The likelihood

that a suitable product is found standard the first control equal to 0,95 and the

second – 0,9. Find the probability that the product test will be recognized as

standard.

4.83 Among 350 mechanisms there are 160 of the first class, 110 second

class, 80 third grade. The probability of marriage arrangements for the first grade is

0.01, for mechanisms of the second grade is 0.02, mechanisms for third grade is

0.04. Taken one mechanism. Find the probability that this mechanism is in order.

4.84 9 white and 5 black balls in the first urn, in the second - three white and

seven black balls. From the second ballot box in the first shifted one ball and then

the urn was taken out from the first one ball. Find the probability that a white ball

was taken.

4.85 The pilot catapulted in the area, 60% of which is covered by forests. The

probability of successful pilot landing in the forest is 0,3 and in treeless areas – 0,9.

Find the probability of successful pilot landing.

4.86 The probability of catching Lyme disease after оn day of hiking in the

Сuуаmаса mountains are estimated at less than 1 in 10000. Yоu feel bad after а day

of hike in the Cuyamaca and decide to take а Lyme disease test. The test is positive.

The test specifications say that in an experiment with 1000 patients with Lyme

disease, 990 tested positive. Moreover when the same test was performed with 1000

patients without Lyme disease, 200 tested positive. What are the chances that you

got Lyme disease.

60

4.87 This рroblem uses Bayes' theorem to combine probabilities as subjective

beliefs with probabilities as relative frequencies. А friend of yours believes she has

а 50% chance of being pregnant. She decides to take а pregnancy test and the test is

positive. Yоu read in the test instructions that out of 100 non-pregnant women, 20%

give false positives. Moreover, out of 100 pregnant women 10% give false

negatives. Help your friend upgrade her beliefs.

The Bernulli formula. Generalizing the discussion of the given example, it is

possible to derive one of the fundamental formulas of probability theory: if events

nAAA ,...,, 21 are independent and each has a probability p, then the probability of

exactly m such events occurring is:

mnmm

nn qpCmP −=)( .

Here, m

nC denotes the number of combinations of n elements taken m at a

time.

For large n, the calculation using this formula becomes difficult. In the

preceding example, let the number of shots equals 100; the problem then becomes

one of finding the probability x that the number of hits lies in the range from 8 to

32. The use of Bernulli formula and the addition theorem gives an accurate, but not

a practically useful, expression for the desired probability:

=

−32

8

100100 8,02,0

m

mmmC .

We can use the Bernulli`s formula for independent events, if probability of

events occurring are:

1) Less then m times:

)1(...)1()0()( −+++= тРРРmР nnn.

2) More than m times:

)(...)2()1()( nРmРmРmР nnn +++++= .

3) No more then m times:

)(...)1()0()( тРРРmР nnn +++= .

4) At least m times:

)(...)1()()( nРmРmРmР nnn ++++= .

61

5) At least once:

)0(1)0( nn ÐmÐ −= .

4.88 Calculate the probability of rolling 4 on a dice exactly 5 times in 25

trials.

Solution. We have the following:

- n = total trials = 25;

- k = total successes = 5;

- n–k = total failures = 20;

- p = 1/6 = 0,167; q = 5/6 = 0,833.

.13053!20!5

!25525 ==C

Therefore, probability (P) will be:

.17844,0833,0167,013053)5,25( 205205525 === qpCP

Thus, the probability is 0,17844. This way, we can calculate the probability

of any event provided we know the number of trials and the probability of the

event occurring in a single trial.

4.89 Kanat is writing an exam of multiple choice questions. It contains a total

of 15 question, each of which has 4 possible answers. What is the probability that

Kanat gets exactly 11 correct answer?

Solution. Given that:

- n = 15;

- p = Probability of success = 1/4 = 0,25;

- q = Probability of failure = 1 – 0,25 = 0,75;

- k = 11.

The probability of exactly 11 correct answers:

0003089,075,025,0)11,15( 4111115 == CP .

4.90 You are taking a 10 question multiple choice test. If each question has

four choices and you guess on each question, what is the probability of getting

exactly 7 questions correct?

Solution. n= 10; k = 7; n – k= 3, p= 0,25 = probability of guessing the correct

answer on a question;

q = 0,75 = probability of guessing the wrong answer on a question:

(7 10 )P correct guesses out of questions =

62

7 7 310 0,25 0,75 0,0031.C= =

4.91 A student takes a multiple choice quiz with 4 possible answers to each of

the 10 questions. If he guesses randomly, find the:

a) probability he scores 7 out of 10;

b) probablility he scores 8 or better;

c) probability he fails (6 or less).

Solution.

a) probability he scores 7 out of 10:

003,075,025,0)7,10( 37710 == CP ;

b) probablility he scores 8 or better:

( 8) (10) (9) (8)P P P P = + + =

10 10 0 9 9 1 8 8 210 10 100,25 0,75 0,25 0,75 0,25 0,75 0,00042.C C C= + + =

c) probability he fails (6 or less):

( 6) 1 [ (10,7) ( 8)]P P P = − + =

7 7 3101 [ 0,25 0,75 0,00042] 1 0,00342 0,9966.C= − + = − =

4.92 Ascar’s free-throw percentage was 0,687. Suppose that shooting free-

throws is a Bernoulli process. If Ascar took 8 free-throws in a certain game that

year, what is the probability that he: makes all 8?

Solution:

8 8 08( ) 0,687 (1 0,687) 0,04962.P A C= − =

4.93 In each of 4 races, the Democrats have a 60% chance of winning.

Assuming that the races are independent of each other, what is the probability that:

(a) The Democrats will win 0 races, 1 race, 2 races, 3 races, or all 4 races; (b) The

Democrats will win at least 1 race; (c) The Democrats will win a majority of the

races.

Solution. X equal the number of races the Democrats win.

a) Using the formula for the binomial distribution:

0 0 4 0 44

4!(4,0) 0,6 0,4 0,0256;

0! 4!P C p q= = =

63

1 1 3 1 34

4!(4,1) 0,6 0,4 0,1536;

1!3!P C p q= = =

;3456,04,06,0!2!2

!4)2,4( 22222

4 === qpCP

;3456,04,06,0!1!3

!4)3,4( 13133

4 === qpCP

.1296,04,06,0!0!4

!4)4,4( 04044

4 === qpCP

b) The Democrats will win at least 1 race:

P(at least 1) = P(X≥1) =1 - P(none) =1 - P(0) = 0,9744.

Or:

P(at least 1) = P(1) + P(2) + P(3) + P(4) = 0,9744.

c) The Democrats will win a majority of the races:

P(Democrats will win a majority) =

= P(X≥3) = P(3) + P(4) = 0,3456 + 0,1296 = 0,4752.

4.94 The probability of hitting the target in one shot is 0,4. It was made 7

shots independent. Find the probability of:

a) two target hits;

b) no more than two target hits.

4.95 Find the probability that in a family with seven children, would be:

a) 4 girls and 3 boys;

b) at least 3 boys (the probability of the birth of a boy is 0,515).

4.96 The probability that a basketball player threw the ball in the basket, is

0.4. Produced 6 shots. Find the probability of:

a) 4 results;

b) at least 2 hits.

4.97 The instrument consists of 6 knots. The probability of failure-free

operation for a time t for each node is equal to 0,7. Find the probability that in time t

refuse:

a) two nodes;

b) no more than two nodes.

64

4.98 Probability of production of high quality parts is 0,7. What is the

probability that among the 10 items:

a) exactly 8 pieces of excellent quality;

b) at least 9 parts of high quality

4.99 10 coin toss. Find the 4 times the probability that fall coat:

a) three times;

b) at least two times.

4.100 The probability of hitting the target in one shot is 0,3. It was made 8

shots independent. Find the probability of:

a) two target hits;

b) no more than two target hits.

4.101 Find the probability that in a family with seven children, would be:

a) 5 girls and 2 boys;

b) at least 2 boys (the probability of the birth of a boy is 0,45).

4.102 The probability that a basketball player threw the ball in the basket, is

0.3. Produced 9 shots. Find the probability of:

a) 5 results;

b) at least 3 hits.

5 Discrete random variables

Definition. Random Variable. If S is a sample space with a probability

measure and X is a real-valued function defined over the elements of S, then X is

called a random variable.

We symbolize random variables by capital letters (e.g., X), and their values

by lower-case letters (e.g., x).

A random variable X is called discrete if it can accept only a finite or

countable range.

Definition (Probability Distribution). If X is a random variable, the function

f(x) whose value is P(X = x) for each value x in the range of X is called the

probability distribution of X.

The following table is called low or series of distributions of a discrete

random variable:

x x1 x2 … xn

p p1 p2 … pn

Thus, it is characterized by the values ...,, 21 xx , which it can accept, and by

the probabilities )( ii xXPp == with which it accepts these values and which should

satisfy to the condition =i

ip 1.

A single-valued mapping of a set ix onto set ip is considered as a function of

probability of a discrete random variable.

65

The graphic representation of the series of the distribution of a discrete

random variable is called the polygon of the distribution (figure 2).

Figure 2

Let X be a random variable. A function )()( xXPxF = is called a

distribution function )(xF of a random variable X. For a distribution function of a

discrete random variable we have:

=xx

i

i

pxF .)(

The value of the distribution function at a point 0x is equal to the probability

that a random variable takes a value less than 0x . In probability theory, a random

variable is completely characterized by its distribution function, i.e. can be

considered as given if its distribution function is given.

Using the distribution function, you can specify the probability that a random

variable falls within a given half-open interval:

)()()( aFbFbXaP −= .

The distribution function )(xF of an arbitrary random variable has the

following properties:

1) .0)(lim,1)(lim ==−→+→

xFxFxx

;

2) )(xF monotonically does not decrease, that is, if equality )()( 21 xFxF if

21 xx ;

3) )(xF is continuous from the left.

Numerical characteristics. Mathematical expectation. The number оf the form

(1) is called the mathematical expectation of a discrete random variable:

1 1 2 2( ) ... ... .n n i i

i

M X p x p x p x p x= + + + + =

and denoted by M(X).

66

The expectation determines the position of the distribution center in the

following sense: if we assume that kp are the masses placed at the points kx of the

real axis, then M(X) is the coordinate of the center of gravity of this system.

Properties of mathematical expectation:

1) The mathematical expectation of the constant C (which can be regarded as

a discrete random variable with one possible value C, which it takes with

probability is equal to this constant:

М(С)=С, C - const.

2) The mathematical expectation of a sum is equal to the sum of mathematical

expectations:

)()()( 2121 XMXMXXM +=+ .

3) The mathematical expectation of a product of constant value by a

random variable is equal to the product of the constant by the mathematical

expectation of a random variable:

)()( XMÑÑXM = .

4) The mathematical expectation of the product of two independent random

variables is equal to the product of their mathematical expectations:

)()()( 2121 XMXMXXM = .

If we have a binomial distribution with parameters n, p:

=

− ==n

k

knkkn npqpkCXM

0

)( .

Dispersion is the mathematical expectation of the square of the difference (

)(XMX − ) and is denoted by D(X), that is:

−=−=k

kk XMXMpXMxXD 22 ))(()]([)( .

To calculate the dispersion, the following formula is often useful:

22 )]([)()( XMXMXD −= .

67

The square root of the dispersion is called the spread, or the standard

deviation, or the mean square deviation, and is denoted by X or )(X :

)(XDX = .

The value or D(X) is the measure of scattering of the distribution with

respect to the mathematical expectation.

Dispersion properties:

1) The dispersion of a constant is zero:

0)( =CD .

2) The dispersion of a product of a constant value by a random variable is

equal to the product of the square of a constant value by the dispersion of the

random variable:

)()( 2 XDCCXD = .

3) The dispersion of the sum of the constant C and the random variable is

equal to the dispersion of the random variable:

)()( XDXCD =+ .

4) The dispersion of the sum of two independent random dispersions is equal

to the sum of the dispersions of these quantities:

)()()( YDXDYXD +=+ .

For Binomial distribution we have:

npqXD =)( .

For Poisson distribution:

== XXD ,)( .

5.1 Given an experiment in which we roll a pair of 4-sided dice, let the

random variable X be the total number of points rolled with two dice.

E.g. X = 5 «picks out» the set {(1,4), (2,3), (3,2), (4,1)}.

Specify the full function denoted by X and determine the probabilities

associated with each value of X.

68

5.2 Assume a balanced coin is flipped three times. Let X be the random

variable denoting the total number of heads obtained:

Outcome Probability

HHH 1/8

HHT 1/8

HTH 1/8

THH 1/8

TTH 1/8

THT 1/8

HTT 1/8

TTT 1/8

Hence:

P(X=0)=1/8; P(X=1)=P(X=2)=3/8; P(X=3)=1/8.

For the probability function defined in the previous example:

x 0 1 2 3

f(x) 1/8 3/8 3/8 1/8

A probability distribution is often represented as a probability histogram. For

the previous example:

f(x)

0,4

0,3

0,2

0,1

0 1 2 3 x

Figure 3

For example, in a raffle, there are 10 000 tickets. The probability of winning

is therefore 00010

1 for each ticket. The prize is worth 4 800$. Hence the expectation

per ticket is 4 800$0,48$

10 000= .

In this example, the expectation can be thought of as the average win per

ticket.

69

This intuition can be formalized as the expected value (or mean) of a random

variable.

5.3 A balanced coin is flipped three times. Let X be the number of heads.

Then the probability distribution of X is:

=

=

=

=

=

38/1

28/3

18/3

08/1

)(

xfor

xfor

xfor

xfor

xf.

The expected value of X is:

.2

3

8

13

8

32

8

31

8

10)()(

4

1

=+++===i

ii xfxXM

The notion of expectation can be generalized to cases in which a function

g(X) is applied to a random variable X.

Theorem. Expected Value of a Function. If X is a random variable and f(x) is

the value of its probability distribution at x, then the expected value of g(X) is:

=

=n

iii xfxgXgM

1

)()())(( .

5.4 Let X be the number of points rolled with a balanced (6-sided) dice. Find

the expected value of X and of 12)( 2 += xXg .

The probability distribution for X is f(x) = 1/6. Therefore:

6

21

6

1)()(

6

1

6

1

=== == i

ii

ii xxfxXM ;

6

94

6

1)12()()())((

6

1

26

1

=+== == ii

ii xxfxgXgM .

Binomial Random Variable. А binomial random variable is а random variable

that counts the number of successes in а sequence of independent Веrnоulli trials

with fixed probabi1ity of success.

А calculating the probabi1ity distribution of Х means calculating the

following probabilities: the probabi1ity of 1 success; Р(X = 1). The probability of 2

successes; Р(X = 2), and so оn. The formula below is derived in the textbook. То

use it, you should know how to compute the binomial coefficients С(m, n).

Probability Distribution of Binomial Random Variable. If Х is the number of

successes in а sequence of n independent Веrnоulli trials, then:

70

knkkn qpCkXP −== )( .

Where

- n - number of trials;

- p - probability of success;

- q - probability of failure pq −=1 .

Example. What is the probability of getting heads exactly twice if уоu fliр а

fair coin 6 times?

We have k = 2 number of successes; n = 6 number of trials; р=0,5 probabi1ity

of success; 5,05,011 =−=−= pq probabi1ity of failure.

Probability of getting 2 heads:

2344,00625,025,0155,05,0)2( 26226 ==== −CXP .

5.5 What is the probability of getting heads exactly 3 times if you flip а fair

coin 6 times?

We have k = 3 number of successes; n = 6 number of trials; р=0,5 probabi1ity

of success; 5,05,011 =−=−= pq probabi1ity of failure.

Probability of getting 3 heads:

3 3 6 36( 3) 0,5 0,5 20 0,125 0,125 0,3125.P X C −= = = =

5.6 The probability of a boy's birth is 0,515. How great is the probability that

among 10 randomly chosen newborns there will be 6 boys?

The assumption of independence can be considered as fulfilled. Thus, for the

desired probability we have:

.2167,0)485,0()515,0()6,10()( 46610 == CPAP

5.7 Let the probability of obtaining a defective product is 0,01. What is the

probability that among a hundred products there will be no more than three

defective ones?

According to the binomial law and the law of addition, we get that:

0 0 100 1 1 99100 100( ) (0,01) (0,99) (0,01) (0,99)P A C C= + +

2 2 98 3 3 97100 100(0,01) (0,99) (0,01) (0,99) 0,9816.C C+ + =

Now that we know how to calculate individual probabilities of the form Р(Х

= х), let us put them аll together to obtain а probability distribution:

5.8 Your nаmе is Asan and you are an expert penalty goal shooter. Your skill

has been improved for the past 10 years, and now you are as good as you will ever

71

bе. Your success rate has been measured at 80%. Thus, p = 0,8 and q=0,2. Yоu take

n = 6 shots оn goal, so the possible values of X (the number of successes) are

0,1,2,3,4,5,6. Here is the probability for each value of X:

0 0 66( 0) 0,8 0,2 1 1 0,000064 0,000064P X C= = = = ;

001536,000032,08,062,08,0)1( 5116 ==== CXP ;

01536,00016,064,0152,08,0)2( 4226 ==== CXP ;

08192,0008,0512,0202,08,0)3( 3336 ==== CXP ;

24576,004,04096,0152,08,0)4( 2446 ==== CXP ;

393216,02,032678,062,08,0)5( 1556 ==== CXP ;

262144,01262144,062,08,0)6( 0666 ==== CXP .

Putting them all together gives the probability distribution for Х:

x 0 1 2 3 4 5 6

Р(Х=х) 0,00006

4

0,00153

6

0,0153

6

0,0819

2

0,2457

6

0,39321

6

0,26214

4

We сan use the probability distribution to find the probability that X is in а

given range bу adding the individual probabilities:

Probability of at least

5 successes

(5 6) (5) (6)

0,393216 0,262144 0,65539

P X P P = + =

= + =

or а 65,5% chance.

Probability of at most

2 successes 01696,001536,0001536,0000064,0

)2()1()0()20(

=++=

=++= PPPXP

or а 1,7% chance.

Probability of at least

3 successes 98304,001696,01

))2()1()0((1)63(

=−=

=++−= PPPXP

i.e. 1-Probability of at most 2

successes.

Since "at most 2

successes" and "аt

least 3 successes" are

соmplеmеntаrу

P=1-0,01696=0,98304.

72

events

We used the previous answer, and that was а easier than adding Р(Х = 3),

P(X = 4), Р(Х = 5) and Р(Х = 6).

Here is the probability distribution again:

x 0 1 2 3 4 5 6

Р(Х=х) 0,00006

4

0,00153

6

0,0153

6

0,0819

2

0,2457

6

0,39321

6

0,26214

4

Poisson distribution. The Poisson distribution, named after French

mathematician Simion Denis Poisson, is а discrete probability distribution that

expresses the probability of а given number of events occurring in а fixed interval

of time and/or space if these events occur with а known average rate and

independently of the time since the last event. The Poisson distribution can also bе

used for the number of events in other specified intervals such as distance, area or

volume.

Introduction to the Poisson distribution. The Poisson distribution is an

appropriate model if the following assumptions are true.

− k is the number of times an event occurs in an interval and k can take

values 0,1,2, ...;

− the occurrence of one event does not affect the probability that а second

event will occur. That is, events occur independently;

− the rate at which events occur is constant. The rate cannot bе higher in

some intervals and lower in other intervals;

− two events cannot occur at exactly the same instant;

− the probability of an event in an interval is proportional to the length of

the interval.

If these conditions are true, then k is а Poisson random variable, and the

distribution of k is а Poisson distribution.

Probabllity of events for а Poisson distribution. An event can occur 0,1,2,…

times in an interval. The average number of events in an interval is designated

(lambda). Lambda is the event rate, also called the rate parameter. The probability

of observing k events in an interval is given bу the equation:

!)(

k

eervalintinevenskP

k −

= .

Where

- is the average number of events per interval;

- е is the number 2,71828… (Euler's number) the base of the natural

logarithms;

- k takes values 0,1,2,…;

73

- k! is the factoria1 of kkkk −−= )1()2(21! .

This equation is the probability mass function (PMF) for а Poisson

distribution.

The Poisson distribution can be used as a good approximation of the

binomial distribution if n is large and p is small. Then, we should take product np

in the capacity of , i.e. np= .

Definition. А discrete random variable Х is said to have а Poisson distribution

with parameter 0 , if, for k=0,1,2, ... the probability mass function of Х is given

bу:

( ) .!

keP X k

k

−

= =

Example. The car traveled 100 000 km. Let X is the number of punctures of

the tire at this distance. Then X can be regarded as a random variable distributed

according to Poisson's law (with a suitable ), that is, the probability of three

punctures of the tire is:

−= eAP!3

)(3

.

5.9 Let the probability of obtaining a defective product is 0,01. What is the

probability that among a hundred products there will be no more than three

defective ones?

We have n=100, p=0,01. Thus 101,0100 === np

,9810,06

1

2

111

1

!3

1

!2

1

!1

1

!0

1)( 1

31

21

11

0

=

+++=

=+++= −−−−

e

eeeeAP

which gives a good match with the exact value, but it is calculated much faster.

5.10 Оn а particular river, overflow floods occur once every 100 years оn

average. Calculate the probability of k=0,1,2,3,4,5 or 6 overflow floods in а 100-

year interval, assuming the Poisson model is appropriate.

As the average event rate is one overflow flood per 100 years, lambda 1= :

1( 100 )

! !

k ke eP k overflow floods in years

k k

− −

= = ;

1 11

( 0 100 ) 0,368;0! 1

k e eP k overflow floods in years

− −

= = = =

74

1 11( 1 100 ) 0,368;

1! 1

k e eP k overflow floods in years

− −

= = = =

2 1 11

( 2 100 ) 0,184.2! 2

e eP k overflow floods in years

− −

= = = =

Тhe table below gives the probability for 0 to 6 overflow floods in а 100 year

period:

k 0 1 2 3 4 5 6

P(k) 0,368 0,368 0,184 0,061 0,015 0,003 0,0005

6 Continuous random variables

A random variable is said to be continuous if its distribution function (the

integral distribution function) can be represented in the form:

−

=x

dttfxF )()( .

The function )(xf is called the distribution density. Since 1)(lim =+→

xFx

, then

the condition:

+

−

=1)( dxxf

should be fulfilled.

Properties of continuous random variables:

1) 0)( xf .

2) ( ) 1.f x dx

+

−

=

For a given probability density, by virtue of the fact that:

)()()( aFbFbXaP −=

and −

=x

dttfxF )()( , the probability that a random variable falls within a given

interval is equal to (figure 4):

75

Figure 4

( ) ( ) ( ) ( ) .

b

a

P a X b F b F a f x dx = − =

The probability P(X = a), i.e. the probability that a continuous random

variable is equal to a given real number, is always 0. Note that equality P(A)=0 does

not imply that A is an impossible event, although P(V)=0, where V is an impossible

event.

Mathematical expectation. This

+

−

= dxxxfXM )()(

is called the mathematical expectation of the random variable X.

The mathematical expectation M(X) gives the position of the center of gravity

of the mass distribution, which is given by the «mass distribution density».

To calculate the dispersion, the following formula is often useful:

+

−

= dxxfxXD )()( 2 .

The quantity =)(XD is called the mean square deviation of the random

variable X

The mathematical expectation, dispersion and mean square deviation in the

continuous case has the same properties 1)-4), which were noted for the discrete

case.

Uniform, exponential, normal distributions. Uniform distribution. A random

variable is called uniformly distributed on ],[ ba if its probability density on the

],[ ba is constant, and outside ],[ ba is equal to 0 (figure 5).

Since:

76

+

−

=1)( dxxf .

Then:

.1

)(ab

xf−

=

Figure 5

Uniform distribution on [a,b]:

+

=−

=b

a

abdx

abxXM

2

1)( ;

−

=−

+−=

b

a

abdx

ab

baxXD

12

)(1

2)(

22

.

Normal distribution (Gaussian distribution). A random variable is said to be

normally distributed if it has a probability density of the following kind:

22 2/)(

2

1)(

axexf −−= , (*)

where a and are the distribution parameters.

Function (*) is a bell-shaped curve (Figure 6).

77

Figure 6

The parameter a is the maximum point through which the symmetry axis

passes, the parameter is the distance from this axis to the inflection point.

If is small, the curve is high and pointed; If is large, it is wide and flat.

Figure 6 shows the normal distribution for a=0 and different .

If the random variable X has a normal distribution with parameters a and ,

then we say that X distributed normally according to law ),,( axN , write

),,( axNX .

Function:

)1,0(,2

1)(

2

=== −

aex x

is called the density of a normalized and centered normal distribution. The

probability density )(x and the corresponding distribution:

Φ(𝑥) = ∫ 𝜑(𝑡)𝑑𝑡.

𝑥

−∞

Is tabulated (Application 2). The function Ф(х) is often called the Gaussian

error integral:

Φ0(𝑥) =1

√2𝜋∫ 𝑒−

𝑡2

2 𝑑𝑡

𝑥

0

;

Φ(𝑥) = Φ0(𝑥) +1

2.

Normal distribution on ),,( axN :

+

−

−− == adxexXM ax 22 2/)(

2

1)(

.

Consequently, the parameter a has the meaning of a mathematical

expectation.

Normal distribution on ),,( axN :

+

−

−− =−= 22/)(2 22

2

1)()(

dxeaxXD ax .

78

Thus, the normal distribution is completely determined by specifying the

mathematical expectation and the standard deviation.

Exponential distribution. A random variable is called exponentially

distributed if it has the following probability density:

𝑓(𝑥) = {𝜆𝑒−𝜆𝑥 , 𝑥 ≥ 0,

0, 𝑥 < 0.

Where is the distribution parameter.

Example. The service life of the light bulb can be viewed with good

approximation as an exponentially distributed value. Figure 7 shows the probability

density of the exponential distribution with 1= :

Figure 7

For exponential distribution:

− ==0

1)(

dxexXM x ;

2

1)(

=XD .

Thus, we have the following table:

The distribution laws of random variables

Uniform ],[ baRX

0, [ , ]

( ) 1, [ , ]

x a b

f xx a b

b a

= −

( )2

a bM X

+=

2( )

( )12

b aD X

−=

If ],[],[ badc , then

79

−

−

=

bx

bxaab

ax

ax

xF

,1

,

,0

)(

( )d c

P c X db a

− =

−

Exponential

=

−

0,0

0,)(

x

xexf

x

−

=

− 0,1

0,0)(

xe

xxF

x

1( )M X

=

2

1( )D X

=

( ) a bP a X b e e − − = −

Normal ),( baNX 2

2

( )

21

( )2

x a

f x e

−−

=

2

21

( )2

x a

t

F x e dt

−

−

−

=

( )M X a= 2( )D X =

( )P X

a

=

− − = −

where 2

21

( )2

x t

Ф x e dt

−

−

= .

6.1 Find the distribution of a discrete random variable X, which has only two

possible values x 1 and x 2 , and x1 <x 2 , knowing the expectation M(X) = 0,24, the

probability of possible value x 1 p 1 =0,6.

Solution: The sum of the probabilities of all possible values of a discrete

random variable is equal to one, so the probability that X takes the value x 2 is equal

to 1 - 0,6 = 0,4. Then the law of the distribution of X is:

Х x 1 x 2

Р 0,6 0,4

To find x 1 and x 2 be two equations, using the known values and formulas

expectation and variance. To do this, write the law of distribution of X 2 :

Х 2 x 2

1 x 2

2

Р 0,6 0,4

Thus:

80

M(X) = 0,6 x 1 + 0,4 x 2 ;

D(X) = 0,6 x 2

1+ 0,4 x 2

2.

Hence we obtain the system:

=−+

=+

24,04,14,06,0

4,14,06,0

22

2

2

1

21

хх

хх.

Solving this system of equations, find two solutions:

x 1 = 1, x 2 = 2 and x 1 = 1,8 and x 2 = 0,8.

According to the problem х 1 <х 2 , so the problem satisfies only the first

solution x 1 =1, x 2 =2. Seeking the law of distribution of a discrete random variable X

is:

Х 1 2

Р 0,6 0,4

6.2 Find the probability of getting a given interval (12,14) normally

distributed random variable, if you know the expectation of a = 10 and standard

deviation = 2.

Solution: we use the formula for the solution:

.)(

−−

−=

аФ

аФXP

Substituting 2,10,14,12 ==== а obtain:

Р(12<Х<14) = Ф(2) - Ф(1).

From the application 2 find the values of the functions:

Ф(2) = 0,4772 and Ф(1) = 0,3413,

then the desired probability P(12<Х<14) = 0,1359.

6.3 A discrete random variable X is set next distribution:

Х 0 10 20 30 40 50

Р 0,05 0,15 0,3 0,25 0,2 0,05

81

Find:

a) distribution function F(x);

b) the math expectation, dispersion.

Solution:

a) for a discrete random variable distribution function F(x) to be for all values

of i, that x i <x according to the formula:

( ) ( ) ( ).

i i

i i

x x x x

F x P X x p P X x =

= = = =

If x ≤ 0, then:

F(x) = P(X < 0) = 0;

If 0 < x ≤ 10, then:

F(x) = P(X = 0) = 0,05;

If 10 <x ≤20, then:

F(x) = P(X = 0) + P(X = 10) = 0,05 + 0,15 = 0,2;

If 20 <x ≤ 30, then:

F(x) = P(X = 0) + P(X = 10) + P(X = 20) = 0,2 + 0.3 = 0,5;

If 30 <x ≤ 40, then:

F(x) = P(X = 0) + P(X = 10) + P(X = 20) + P(X = 30) = 0,5 + 0,25 = 0.75;

If 40 < x ≤ 50, then:

F(x) = P(X = 0) + P(X = 10) + P(X = 20) + P(X = 30) +

+ P(X = 40) =0,75 + 0,2 = 0,95;

If x < 50, then:

F(x) = P(X = 0) + P(X = 10) + P(X = 20) + P(X = 30) +

+ P(X = 40) + P(X = 50) = 0,95 + 0,05 = 1.

In this way:

82

0, 0

0.05, 0 10

0.2, 10 20

( ) 0.5, 20 30

0.75, 30 40

0.95, 40 50

1, 50,

x

x

x

F x x

x

x

x

=

;

b) we find the math expectation and the dispersion.

The math expectation for discrete random variable is:

( ) 0 0.15 10 0.15 20 0.3 30 0.25i i

i

М Х х p= = + + + +

.5.2505.0502.040 =++

The dispersion of the random variable X is obtained by the formula:

D(X) = 22 )()( ХМХМ − .

We have:

2 2 2 2( ) 0 0,05 10 0,15 20 0,3 30 0,25D X = + + + +

2 240 0,2 50 0,05 805.+ + =

6.4 Find the distribution of a discrete random variable X, which has only two

possible values 𝑥1 and 𝑥2, 𝑥1 < 𝑥2, if you know the math expectation of M(X)=3,1

the dispersion of D(X) = 0,09 and the probability p1=0,9 possible value x1.

6.5 40% of the production plant is the second grade products. Find the

probability that at the batch of 150 items it would be 63 items the second grade.

6.6 Probability of production of defective parts is 0,02. Find the most

probable number of defective components in a batch of 160 parts and the

corresponding probability.

6.7 Has 75 boxes of the same type of parts. The probability that one randomly

selected box has non-standard details equal to 0,25. Find the probability that the 15

boxes will be non-standard parts.

6.8 The probability that the dealer will sell the security, equal to 0,6. Find the

probability that out of 150 securities him he would be able to sell of hitting a target

on one shot is equal to 0,8. Find the probability of 27 misses on 100 shots.

6.9 Dice tossed 360 times. Find the probability that the number of deletions

six points will be concluded between 50 and 65.

83

6.10 The probability that the family has a refrigerator, equal to 0,9. Find the

probability that out of 400 families from 355 to 375 have a fridge.

6.11 The probability of hitting the target is 0,5. Find the probability that the

number of hits 250 shots will be from 120 to 140.

6.12 From an urn containing four white and six black balls, one ball at

random is removed and recycled. The test is repeated 100 times. Find probability

that a white ball will be from 35 to 50 times.

6.13 Random variable X is given normally of distributed with mathematical

expectation a=13, standard deviation 5= . Find:

1) distribution density and distribution function;

2)mathematical expectation, dispersion;

3) the probability of getting at the interval (6;18).

6.14 Discrete random variable X is given by law of distribution. Find:

a) distribution function F(x) and construct the graph;

b) mathematical expectation, dispersion, standard deviation, mode:

X -1 0 2 3 7 8

p 0,21 0,16 0,14 0,1 0,2 0,13

6.15 A random variable is given by the distribution function:

2

0, 1,

( 1)( ) , 1 3,

4

1, 3.

x

xF x x

x

−=

Find:

1) The distribution density;

2) Mathematical expectation;

3) The probability of getting at the interval (0;2);

4) Mode;

5) Median.

6.16 A random variable has a distribution density:

0, 1, 4,

( ) 1, 1 4.

3

x x

f xx

=

Find:

1) The distribution density;

2) Mathematical expectation;

3) The probability of getting at the interval (0;2);

84

4) Mode;

5) Median.

6.17 The random variable is uniformly distributed on the interval [2, 6]. Find:

1) The distribution density, distribution function;

2) Mathematical expectation;

3) The probability of getting at the interval (1;2).

6.18 The random variable obeys the exponential distribution with parameter

1,0= . Find:

1) The distribution density;

2) Distribution function;

3) Mathematical expectation.

6.19 Discrete random variable X is given by a series of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

Х 2 4 5 6 8

р 0,1 0,2 0,3 0,1 0,3

6.20 Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

X -3 1 4 5 7 9

p 0,15 0,2 0,2 0,1 0,15 0,2

6.21 Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

X 1 1 2 3 4

p 0,2 0,3 0,1 0,2 0,2

6.22 Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

X -2 0 2 4

p 0,35 0,15 0,23 0,27

6.23 Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

85

X 0 1 2 3

p 0,21 0,16 0,34 0,29

6.24 Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

X -1 0 2 3 7 8

p 0,2 0,1 0,4 0,1 0,1 0,1

6.25. Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

X -3 0 4

p 0,37 0,27 0,36

6.26 Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

X 2 4 5 6

p 0,25 0,16 0,34 0,25

6.27 Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

X 2 3 7 8

p 0,14 0,26 0,2 0,4

6.28 Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

X -2 0 2

p 0,2 0,5 0,3

6.29 Discrete random variable X is given by law of distribution. Find:

1) Distribution function F(x) and construct the graph;

2) Mathematical expectation, dispersion, standard deviation, mode:

X -1 0 2 3 7

p 0,27 0,17 0,13 0,1 0,33

86

Limits theorems of probability theory. Law of large numbers. A sequence

}{ nX of random variables is called convergent in probability to a random variable X

if for any 0 the equality:

1)(lim =−

→XXP n

n

must be satisfied.

It is said that the sequence of random variables ,..., 21 XX obeys the weak law

of large numbers if for any 0 the equality:

1)(11

lim11

=

−

==→

n

kk

n

kk

nXM

nX

nP

is satisfied.

In other words, if:

==

−=n

kk

n

kkn XM

nX

nZ

11

)(11

converges to 0 «in probability».

To derive the weak law of large numbers, Chebyshev's inequality is

important. Let X be a random variable having finite variance, then for any 0 the

inequality:

2

)())((

XDXMXP −

is satisfied.

Chebyshev's theorem. If ,..., 21 XX is a sequence of pairwise independent

random variables with equal mathematical expectations aXMXM === ...)()( 21 , and

)( kXD are uniformly bounded, i.e. CXD k )( for each k, then for any 0 :

11

lim1

=

−

=→

n

kk

naX

nP .

De Moivre-Laplace`s local theorem. If the probability of occurrence of an

event p in each test and the constant is not 0 or 1, and n number of trials increases

indefinitely, the probability of occurrence an event at exactly k times in tests is

approximated:

87

)(1

)( õnpq

kPn ,

where 2/2

2

1)( xeõ −=

,

npq

npkõ

−= , pq −=1 , )(õ is function values given by

Application 1.

Properties of )(õ :

1) )(õ is even function: )()( хх =− .

2) if 4x , then 0)( =x .

In other words the binomially distributed random variable is asymptotically

distributed normally with the parameters а=пр and npq= .

6.30 Let the probability of appearance in the production of a defective part be

0,005. How great is the probability that among the 10 000 items 40 will be

defective? So, it should be determined ),( knP for n=10 000, k=40, p=0,005. We

have:

)(1

)( õnpq

kPn

and

;05,7=npq .42,1−=−

npq

npk .

From Application 1 we find .1456,0)42,1( = Thus:

0206,005,7

1456,0)40,00010( =P .

De Moivre-Laplace`s integral theorem. If the probability of occurrence of an

event p is constant ( 10 p ) and if events of n infinitely increases, then probability

is written as )( 21 kmkP or );( 21 kkP :

)()(2

1),( 12

2/

21

2

1

2

хФхФdtеkkРх

х

t

n −= −

or

𝑃(𝑘1; 𝑘2) = Φ(𝑥2) − Φ(𝑥1).

Where:

88

𝑥1 =𝑘1−𝑛𝑝

√𝑛𝑝𝑞, 𝑥2 =

𝑘2−𝑛𝑝

√𝑛𝑝𝑞.

Ф(х) is supporting function of Laplace function, this function values are

given by Application 2.

Properties of Ф(х):

1) Ф(х) is odd function: Ф(-х)=-Ф(х).

2) if 5x , then Ф(х)=0,5.

The sufficient accuracy of the formula is ensured when 15npq .

6.31 Let the probability of appearance of a defective part in the production be

0,005. How great is the probability that among the 10 000 items 40 will be

defective?

So, it should be determined ),( knP for n=10 000, k=40, p=0,005. We have:

2

2

1

2

1),(

−−

npq

npk

enpq

knP

.

Then:

;05,7=npq .42,1−=−

npq

npk

Consequently:

( )242,12

1

205,7

1)40,00010(

−− eP

.

From Application 1 we find .1456,0)42,1( = . Thus:

.0206,005,7

1456,0)40,00010( =P .

6.32 Let there is a situation described in task 210. We are looking for the

probability that in a box with 10 000 parts there are not more than 70 defective

ones:

50 20( 70)

49,75 49,75

X npP X P

npq

− − = =

22,84

/2

7,09

17,09 2,84

2

xX npP e dx

npq

−

−

−= − =

89

0 0(2,84) ( 7.09).Ф Ф= − −

Since 0 0( ) ( )Ф x Ф x− = − , then:

0 0( 70) (2,84) (7,09)P X Ф Ф = + .

From the Application 2 we find 0(2,84) 0,4977.Ф = There is no 0(7,09)Ф in the

table, since it differs from 0,5 a little bit. Thus, 9977,0)70( XP .

6.33 Assume that a measurement is made and X is a random measurement

error. A random variable X appears as a result of the additive superposition of a

large number of factors (that do not depend on each other) generating errors. Each

of these factors has a small effect on the error. Thus, the value of X can be assumed

to be distributed normally.

6.34 Let X be the length of a birch leaf randomly selected from a certain set

of torn leaves. Then X is a random variable obtained by imposing of many small

factors that do not depend on each other. Therefore, a normal distribution can be

adopted for X.

6.35 The shop produces 75% production premium. Find the probability that a

run of 160 products will be 125 premium products.

Solution: Denoted by A events of randomly selected product premium. By

condition:

( ) 25.01;75.0,125,160 =−===== pqAPpmn .

Finding:

𝑛𝑝 = 160 • 0.75 = 120,

√𝑛𝑝𝑞 = √120 • 0.75 = √30 ≈ 5.477,

125 120 50.91.

5.477 5.477

m npx

npq

− −= = =

From the Application 1 we find: ( ) 2637,091,0 = . Applying local Laplace

formula we obtain the desired probability:

( )( )

0481,0477,5

2637,0125160 ==

npq

xP

.

6.36 Dice is thrown 144 times. Find the probability of getting the edge with

six points from 20 to 25 times.

90

Solution: denoted by A event of getting the edge with six points in a single

throw of the dice. Find the probability of the event A. By condition

( ) 1 2

1 5144, , , 20, 25

6 6n p P A q m m= = = = = = .

Perform the necessary calculations

1144* 24,

6np = =

524* 20 4.427

6npq = = = .

We compute 1x and 2x by formula:

11

20 240.89;

4.472

m npx

npq

− −= = = −

22

25 240.22.

4.472

m npx

npq

− −= = =

We apply the Laplace integral formula, the values of the Laplace found from

Applicathion 2:

( ) ( ) ( ) ( ) ( )144 2 120;25 0,22 0,89P Ф х Ф х Ф Ф − = − − =

( ) ( )0,22 0,89 0,0871 0,3133 0,4004Ф Ф= + = + = .

6.37 40% of the production plant is the second grade products. Find the

probability that it would be 63 items of the second grade out the party of 150 items.

6.38 Probability of production of defective parts is 0,02. Find the most

probable number of defective components in a batch of 160 parts and the

corresponding probability.

6.39 There are 75 boxes of the same type of details. The probability that one

randomly selected box has non-standard details is equal to 0,25. Find the probability

that 15 boxes will have non-standard parts.

6.40 The probability of hitting a target on one shot is equal to 0,8. Find the

probability of 27 misses on 100 shots.

6.41 Dice is planted 180 times. Find the probability that the same figure will

drop from 5 to 40 times.

6.42 It is known that 70% of the production plant is the highest grade

products. Find the probability that will be 75 premium products the batch of 100

items.

6.43 Probability of hitting targets by shooter in one shot is 0,7. Done 50 shots.

Find the probability of 30 hits.

91

6.44 Dice is tossed 72 times. Find the probability that the six points will drop

15 times.

6.45 Throw coin 100 times. Find the probability that the head will fall 45

times.

6.46 Planted 400 trees. Find the probability that exactly the trees do not take

root 90 if the likelihood that an individual will get accustomed tree equals 0,8.

6.47 Probability of target defeat by shooter in one shot is equal to 0,85. Find

the probability that the shooter will hit the target 175 times in 200 shots.

6.48 Chance of white mushroom among others is 0,25. What is the

probability that will be 25 white mushrooms among 80?

6.49 Dice is tossed 360 times. Find the probability that the number of

deletions six points will be concluded between 50 and 65.

6.50 The probability that the family has a refrigerator, is equal to 0,9. Find the

probability that out of 400 families from 355 to 375 have a refrigerator.

6.51 The probability of hitting the target is 0,5. Find the probability that the

number of hits will be from 120 to 140.

6.52 From an urn containing four white and six black balls, one ball at

random is removed and recycled. The test is repeated 100 times. Find probability

that a white ball will be chosen from 35 to 50 times.

6.53 From an urn containing 3 white and 3 black balls, two ball at random

are removed and recycled. The test is repeated 100 times. Find probability that a

white ball will be chosen from 16 to 30 times.

6.54 The probability of having a boy is 0,515. Find the likelihood that infants

will include from 200 100 to 115 boys.

6.55 Assume that a measurement is made and X is a random measurement

error. A random variable X appears as a result of the additive superposition of a

large number of factors (that do not depend on each other) generating errors. Each

of these factors has a small effect on the error. Thus, the value of X can be assumed

to be distributed normally.

6.56 Let X be the length of a birch leaf randomly selected from a certain set

of torn leaves. Then X is a random variable obtained by imposing of many small

factors that do not depend on each other. Therefore, a normal distribution can be

adopted for X.

6.57 Chance of white mushroom among others is 0,3. What is the probability

that will be 40 white mushrooms among 90?

92

Bibliography

1 Гмурман В.Е. Теория вероятностей и математи-ческая статистика.

Учебное пособие для вузов. – М.: Высшая школа, 2013. – 279 с.

2 Khasseinov K. Canons of mathematics. – M.: Nauka. 2007. – 591 p.

3 Письменный Д. Конспект лекций по теории вероятностей и

математической статистике, случайные процессы. – М.: Айрис-Пресс, 2006. –

288 с.

4 Кремер Н.Ш. Теория вероятностей и математическая статистика. 3-е

издание, дополненное и переработанное. – М.: ЮНИТИ-ДАНА, 2010. – 551 с.

5 Гмурман В.Е. Руководство к решению задач по теории вероятностей

и математической статистике. – М.: Высшая школа, 2003. – 400 с.

6 General course of higher mathematics for economists. Textbook edited by

V. I. Ermakov. – M.: INFRA, 2010. – 656 c.

7 Kremer N.Sh. Higher mathematics for economists: Textbook for higher

educational institutions. – M., 1997.

8 Kremer N.Sh. Theory of probabilities and Mathematical Statistics. – M.:

UNITY, 2002.

9 Bugrov Ya. S., Nikolskii S. M. Calculus. – M.: Nauka, 1988.

10 Panteleyev V.P. Probability and Statistics in problems. – Murmansk:

MSTU, 2008.

Contents

Introduction ………………………………………………………………... 3

1 Elements of combinatorics ………………………………………………. 3

2 Algebra of events ………………………………………………………... 17

3 Direct calculation of probabilities ……………………………………….. 20

4 Algebra of probabilities …………………………………………………. 27

5 Discrete random variables ………………………………………………. 64

6 Continuous random variables …………………………………………… 74

Bibliography ………………………………………………………………. 92

Vassilina Gulmira Kazhymuratovna

THE PROBABILITY THEORY

IN EXAMPLES AND TASKS

Manual for high school

Editor M.D. Kurmanbekova

Signed for print ___ ___________ 2019.

Circulation 500 copies. Form 60 х 84 1/16

Offset paper of No. 2

Printed sheet 6,0. Order of No. ___

Price 3000 tenge.

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