Post on 20-Apr-2023
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Chapter 2
Probability & Expectation 2.1 Introduction
The numerical measure of certainty of an event is called probability. The probability of
any event lies between 0 and 1. Probability of a sure event is 1 while that of an
impossible event is 0.
Sample Space: The set of all possible outcomes associated with an experiment is called
sample space. For example while tossing a coin, the sample space is π = π»,π and while
tossing two coins π = π»π»,π»π,ππ»,ππ , whereas while rolling a die π = 1,2,3,4,5,6 .
Event: An event is the subset of a sample space. For example getting an odd number
while rolling a die is an event πΈ = 1,3,5 .
Mutually Exclusive Events: Two or more events
πΈ1,πΈ2,β¦β¦ . ,πΈπ in a sample space are mutually
exclusive if they have no point in common i.e.
if πΈ1 β© πΈ2 β© β¦β© πΈπ = π . Getting an odd number
and getting an even number while rolling a die are
two mutually exclusive events.
Exhaustive Events: Two or more events are called exhaustive events if at least one of
them occurs when an experiment is performed. For example while tossing a coin; there
are two exhaustive cases head and tail. Also while rolling a die; getting an odd number
and getting an even number are two exhaustive events. Whereas getting a number less
than two and getting a number greater than two, in case of rolling a die are non-
exhaustive events.
Mutually Exclusive and Exhaustive Events:
If events πΈ1,πΈ2 ,β¦ ,πΈπ in a sample space are
mutually exclusive and exhaustive then
π(πΈ1) + π πΈ2 + β¦+ π(πΈπ) = 1
Equally Likely Events: Two or more events are
equally likely if they have same probability of occurrence. Getting an odd number and
getting an even number while rolling an unbiased dice are two mutually exclusive and
equally likely events.
Independent Events: Two events are said to be independent if the occurrence or non β
occurrence of one does not affect the probability of occurrence of the other.
Mathematically the events πΈ1 and πΈ2 are independent if and only if
π(πΈ1 β© πΈ2) = π(πΈ1) Γ π πΈ2
If two events are independent; they cannot be mutually exclusive and vice-versa.
Mutually Exclusive & Exhaustive Events
Mutually Exclusive Events
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2.2 Mathematical Definition of Probability
If a trial results in π exhaustive, mutually exclusive and equally likely events and π of
them are favorable to happening of an event πΈ; then probability of happening of πΈ is
given by: π πΈ = Favorable number of cases
Exaustive number of cases=
π(πΈ)
π(π)=
π
π
Example1 Give an example of two events which are mutually exclusive but not
exhaustive.
Solution: In an experiment of tossing two coins,
Let πΈ1: Getting two heads
πΈ2: Getting two tails
Also π = π»π»,π»π,ππ»,ππ
πΈ1 = π»π» and πΈ2 = ππ
β΄πΈ1 β© πΈ2 = π but π πΈ1 + π(πΈ2) β 1
Hence the events πΈ1ππππΈ2 are mutually exclusive but not exhaustive.
Example2 Give an example of two events in an experiment of tossing two coins, which
are mutually exclusive and exhaustive.
Solution: In an experiment of tossing two coins, let
πΈ1: Getting at least one head
πΈ2: Getting two tails
Now S= π»π»,π»π,ππ»,ππ
πΈ1 = π»π»,π»π,ππ» andπΈ2 = ππ
β΄πΈ1 β© πΈ2 = π and π πΈ1 + π πΈ2 = 1
Hence the events πΈ1ππππΈ2 are mutually exclusive and exhaustive.
Example 3 An urn contains 5 white, 6 red and 4 blackballs. Two balls are drawn at
random. Find the probability that both are red. Also find the probability of one white and
one black ball.
Solution: White Balls: 5, Red Balls: 6, Black Balls: 4
Let πΈ: Both balls are red
π πΈ β‘ π red ball and π red ball
βπ πΈ = 6
15Γ
5
14 =
1
7
Let πΉ: One white and one black ball are drawn
π πΉ β‘ π white ball andπ black ball or π black ball andπ white ball
βπ πΉ = 5
15Γ
4
14+
4
15Γ
5
14 =
4
21
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Note: Questions in which replacement is not allowed can be attempted in a better manner
using combinations.
Using combinations, π πΈ = 6πΆ2
15πΆ2
= 6Γ5
15Γ14=
1
7
π πΉ = 5πΆ1 Γ4πΆ1
15πΆ2
= 5Γ4Γ2
15Γ14=
4
21
Example4 An urn contains 4 white, 5 red and 6 black balls. Three balls are drawn at
random. Find the probability that balls are white, red and black.
Solution: White Balls: 4, Red Balls: 5, Black Balls: 6
LetπΈ: Balls are white, red and black
π πΈ = 4πΆ1 Γ5πΆ1 Γ6πΆ1
15πΆ3
= 4Γ5Γ6Γ3!
15Γ14Γ13=
24
91
Example5 Four cards are drawn without replacement from a well shuffled pack of 52
cards. Find the probability that
(i) All cards are spades
(ii) There are two spades and two hearts
(iii) All cards are black
Also compute the probabilities if four cards are drawn with replacement.
Solution: LetπΈ1: All cards are spades
πΈ2: There are two spades and two hearts
πΈ3: All cards are black
π πΈ1 = 13πΆ4
52πΆ4
= 13Γ12Γ11Γ10Γ4!
52Γ51Γ50Γ49Γ4!=
11
4165
π πΈ2 = 13πΆ2 Γ13πΆ2
52πΆ4
=13Γ12Γ13Γ12Γ4!
52Γ51Γ50Γ49Γ2!Γ2!=
468
20825
π πΈ3 = 26πΆ4
52πΆ4
=26Γ25Γ24Γ23Γ4!
52Γ51Γ50Γ49Γ4!=
46
833
Again if the four cards are drawn with replacement
π πΈ1 = 13
52Γ
13
52Γ
13
52Γ
13
52=
1
256
π πΈ2 = 13
52Γ
13
52Γ
13
52Γ
13
52Γ 6 =
3
128
(β΅ Favorable events are SSHH or HHSS or SHSH or SHHS or HSHS or HSSH where S
denotes Spade and H denotes Heart)
π πΈ3 = 26
52Γ
26
52Γ
26
52Γ
26
52=
1
16
Example6 A pair of dice is rolled. What is the probability of sum 7?
Solution: Let E: Getting a sum 7, when a pair of dice is rolled
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S= 1,1 , 1,2 ,β¦ , 1,6 , 2,1 , 2,2 , . . . , 2,6 , 3,1 , 3,2 ,β¦ , 3,6 , 4,1 , 4,2 ,β¦ , 4,6 , 5,1 , 5,2 ,β¦ , 5,6 , 6,1 , 6,2 ,β¦ , 6,6
πΈ = 1,6 , 2,5 , 3,4 , 4,3 , 5,2 , (6,1)
P πΈ = π(πΈ)
π(π)=
6
36=
1
6
Example7 Only 3 events A, B, C can happen. Given that chance of A is one-third that of
B and odds against C are 2:1, find odds in favor of A.
Solution: Given π π΄ + π π΅ + π πΆ = 1 β¦β
Also π π΄ = 1
3π π΅ β π π΅ = 3 π(π΄) β¦β‘
And π πΆ = 1
3 β¦β’
Using β‘, β’ in β
β π π΄ + 3π π΄ +1
3= 1
β4π π΄ = 2
3
βπ π΄ = 1
6 and π π΄ =
5
6
Hence odds in favour of A are 1:5
Example 8 A problem in mathematics is given to three students π΄ , π΅and πΆ whose
chances of solving are 2
3 ,
1
2 and
1
3 respectively. What is the probability that the problem
will be solved?
Solution: Probability of π΄ solving the problem π π΄ = 2
3 β π π΄ =
1
3
Probability of π΅ solving the problem π π΅ = 1
2 β π π΅ =
1
2
Also probability of πΆ solving the problem π πΆ = 1
3 β π πΆ =
2
3
Now probability that π΄, π΅and πΆ do not solve the problem is 1
3Γ
1
2Γ
2
3=
1
9
β΄ The probability that the problem is solved = 1 β (Probability that problem is not
solved) = 1 β1
9=
8
9
Example 9 A bag contains 50 tickets numbered from 1 to 50, out of which 5 are drawn at
random and arranged in ascending order (π‘1 < π‘2 < π‘3 < π‘4 < π‘5). Find the probability
of π‘4 carrying the number 45.
Solution: Exhaustive number of cases = 50πΆ5.
To follow the given pattern, three tickets π‘1 , π‘2 , π‘3 must be drawn out of tickets
numbered from 1 to 44 with favorable number of cases 44πΆ3 , then π‘4 is drawn bearing
number 25 which has only one favourable case 1πΆ1, and then π‘5 is drawn out of remaining
5 tickets with favourable number of cases 5πΆ1.
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β΄ Required probability is 44πΆ3 Γ1πΆ1Γ5πΆ1
50πΆ5
= 44Γ43Γ42Γ5Γ5!
50Γ49Γ48Γ47Γ46Γ3!= 0.03
Example 10 π΄ has two shares in lottery in which there is 2 prizes and 3 blanks; π΅ has
three shares in a lottery in which there are 3 prizes and 6 blanks. Compare the probability
of π΄β²π success to that of π΅β²π success.
Solution: Probability of π΄ getting not getting a prize in two shares =3πΆ2
5πΆ2
=3Γ2!
5Γ4=
3
10
β΄ Probability of π΄ getting a prize=7
10
Probability of π΅ not getting a prize in three shares =6πΆ3
9πΆ3
=6Γ5Γ4
9Γ8Γ7=
5
21
β΄ Probability of π΅ getting a prize=16
21
Hence the probability of π΄β²π success to that of π΅β²π success =7
10:
16
21
= 147: 160
Example 11 What is the probability that a leap year selected at random will have 53
Mondays.
Solution: A leap year has 366 days, having 52 full weeks and 2 extra days.
π = { Monday, Tuesday , Tuesday, Wednesday , Wednesday, Thursday ,
(Thursday, Friday), (Friday, Saturday), (Saturday, Sunday),
(Sunday, Monday)}
Let πΈ: The leap year selected will have 53 Mondays.
πΈ = Monday, Tuesday , (Sunday, Monday)
β΄ π πΈ =π(πΈ)
π(π)=
2
7
Example 12 π΄ and π΅ alternatively throw a die until one gets a success and wins the
game, where success is defined as getting a six. What are there respective probabilities of
winning if π΄ takes the first trial?
Solution: If success ( π ) is getting a six, thenπ π =1
6, and if πΉ denotes failure,
then π πΉ =5
6.
Now π΄ can win in 1st, 3
rd, 5
th , β― trials, i.e. by getting π or πΉπΉπ or πΉπΉπΉπΉπ, β―
β΄ π΄βs probability of winning π(π΄) =1
6+
5
6.
5
6.
1
6+
5
6.
5
6.
5
6.
5
6.
1
6+ β―
This is a G.P. with π =1
6 and π =
5
6.
5
6=
25
36
β΄ π π΄ =π
1βπ=
1
6
1β25
36
=6
11 , π π΅ = 1 β
6
11=
5
11
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2.3 Addition Law of Probability
Statement: If π΄ and π΅ be any two events, then
P π΄ βͺ π΅ = π π΄ + π π΅ β π π΄ β© π΅
Proof: π΄ βͺ π΅ = π΄ βͺ π΅ β© π΄πΆ
β P π΄ βͺ π΅ = π(π΄ βͺ π΅ β© π΄πΆ )
= π π΄ + π π΅ β© π΄πΆ
(β΅Both π΄ and π΅ β© π΄πΆ are disjoint)
= π π΄ + π π΅ β© π΄πΆ + π π΄ β© π΅ β π π΄ β© π΅
= π π΄ + π[ π΅ β© π΄πΆ βͺ π΄ β© π΅ ] β π π΄ β© π΅
= π π΄ + π π΅ β π π΄ β© π΅
β΅ π΅ β© π΄πΆ βͺ π΄ β© π΅ = π΅
Hence proved
If A and B are mutually exclusive events, π΄ β© π΅ = π
β P π΄ βͺ π΅ = π π΄ + π π΅
If π΄, π΅ , πΆ are 3 events, then
π π΄ βͺ π΅ βͺ πΆ = π π΄ + π π΅ + π πΆ β π π΄ β© π΅ β π π΅ β© πΆ β
π πΆ β© π΄ +P π΄ β© π΅ β© πΆ
Example13 Find the probability of getting a spade or an ace when a card is drawn from a
well shuffled pack of 52 cards.
Solution: Let A: Getting a spade, π π΄ =13
52
B: Getting an ace, π π΅ =4
52
π΄ β© π΅: Getting an ace of spade, π π΄ β© π΅ =1
52
Now P π΄ βͺ π΅ = π π΄ + π π΅ β π π΄ β© π΅
= 13
52+
4
52β
1
52 =
16
52 =
4
13
Example14 Find the probability of getting neither heart nor king when a card is drawn
from a well shuffled pack of 52 cards.
Solution: Letπ΄: Getting a card of heart, π π΄ =13
52
π΅: Getting a card of king, π π΅ =4
52
π΄ β© π΅: Getting a king of heart, π π΄ β© π΅ =1
52
Now P π΄ βͺ π΅ = π π΄ + π π΅ β π π΄ β© π΅
= 13
52+
4
52β
1
52 =
16
52 =
4
13
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Again Probability of neither heart nor king is given by π(π΄πΆ β© π΅πΆ)
π(π΄πΆ β© π΅πΆ) = π(π΄ βͺ π΅)πΆ = 1 β π π΄ βͺ π΅
= 1 β4
13 =
9
13
Example15 Three newspapers A, B, C are published in a city and a survey on readers
reveals the following information:
25% read A, 30% read B, 20% read C, 10% read both A and B, 5% read both A and C,
8% read both B and C, 3% read all three newspapers. For a person chosen at random,
find the probability that he reads none of the newspapers.
Solution: π π΄ =25
100, π π΅ =
30
100, π πΆ =
20
100 , π π΄ β© π΅ =
10
100,
π π΄ β© πΆ =5
100, π π΅ β© πΆ =
8
100 , P π΄ β© π΅ β© πΆ =
3
100
Now π π΄ βͺ π΅ βͺ πΆ = π π΄ + π π΅ + π πΆ β π π΄ β© π΅ β π π΅ β© πΆ β
π πΆ β© π΄ +P π΄ β© π΅ β© πΆ
βP π΄ βͺ π΅ βͺ πΆ =25
100+
30
100 +
20
100β
10
100β
8
100β
5
100 +
3
100
= 55
100 =
11
20
βπ(π΄ βͺ π΅ βͺ πΆ)πΆ = 1 β π π΄ βͺ π΅ βͺ πΆ
= 1 β11
20=
9
20
Example 16 Discuss and comment on following:
π π΄ = 1
4, π π΅ =
1
3, π πΆ =
2
3 are probabilities of three mutually exclusive events
π΄, π΅ , πΆ.
Solution: Since π΄, π΅ , πΆ are mutually exclusive events,
P π΄ βͺ π΅ βͺ πΆ = π π΄ + π π΅ + π πΆ
βP π΄ βͺ π΅ βͺ πΆ = 1
4+
1
3+
2
3 =
5
4> 1
Which is not possible, hence it is a false statement.
2.4 Conditional Probability
The probability of occurrence of an event π΄, when event π΅ has already occurred is called
conditional probability of π΄ and is denoted by π π΄|π΅ .
If π π΄|π΅ = π(π΄), then event π΄ is said to be independent of event π΅.
2.4.1 Multiplicative Law of Probability
Statement: The probability of simultaneous occurrence of two events is equal to the
probability of event multiplied by conditional probability of the other, i.e. for two events
π΄ and π΅, π π΄ β© π΅ = π(π΄). π π΅|π΄
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Proof: Suppose a trial results in π exhaustive, mutually exclusive and equally likely
outcomes, π of them being favourable to the happening of event π΄,
then π π΄ = π
π β―β
Out of π outcomes, favorable to the happening of event π΄ , let π1 be favourable to
happening of event π΅
β΄ Conditional probability of π΅ given that π΄ has already occurred is
π π΅|π΄ =π1
π β―β‘
Again out of π exhaustive, mutually exclusive and equally likely outcomes, π1 are
favorable to happening of π΄ and π΅
β΄ Probability of simultaneous occurrence of π΄ and π΅
= π π΄ β© π΅ = π1
π
= π1
π. π
π =
π
π. π1
π = π(π΄). π π΅|π΄ Using β and β‘
β΄π π΄ β© π΅ = π(π΄). π π΅
π΄
π π΄ β© π΅ is also written as π π΄π΅ Interchanging π΄ and π΅, π π΅π΄ = π(π΅). π π΄|π΅
or π π΄π΅ = π(π΅). π π΄|π΅ β΅π΄ β© π΅ = π΅ β© π΄
If π΄ and π΅ are independent events, then π π΅|π΄ = π(π΅)
β΄ π π΄π΅ = π(π΄). π(π΅) , if events π΄ and π΅ are independent
Formulae for conditional probability:
π π΅|π΄ = π(π΄β©π΅)
π(π΄)
π π΄|π΅ = π(π΄β©π΅)
π(π΅)
Example17 An unbiased coin is tossed thrice. In which of the following cases events A
and B are independent
(i) π΄: The first throw results in a tail
π΅: The last throw results in a head
(ii) π΄: The number of tails is two
π΅: The last throw results in a tail
Solution: π = π»π»π»,π»π»π,π»ππ»,π»ππ,πππ,πππ»,ππ»π,ππ»π»
(i) π΄ = πππ,πππ»,ππ»π,ππ»π» , π π΄ =4
8 =
1
2
π΅ = HHH, HTH, TTH, THH , π π΅ =4
8 =
1
2
π΄ β© π΅ = {πππ»,ππ»π»} , π π΄ β© π΅ =2
8=
1
4
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Now π π΄ β© π΅ =1
4, and π(π΄). π π΅ =
1
4
β΄π π΄ β© π΅ = π(π΄). π(π΅)
Hence events π΄ and π΅ are independent.
(ii) π΄ = HTT, THT, TTH , π π΄ =3
8
π΅ = HHT, HTT, TTT, THT , π π΅ =4
8 =
1
2
π΄ β© π΅ = {HTT, THT} , π π΄ β© π΅ =2
8=
1
4
Now π π΄ β© π΅ =1
4, andπ(π΄). π π΅ =
3
16
β΄π π΄ β© π΅ β π(π΄). π(π΅)
Hence the events π΄ and π΅ are not independent.
Example18 If π π΄ = 0.3, π π΅ = 0.5 and π π΄|π΅ = 0.4
Find (i) π(π΄ β© π΅) , (ii) π΅|π΄ , (iii) π(π΄πΆ βͺ π΅πΆ)
Solution: (i) π π΄ β© π΅ = π π΅ .π π΄
π΅ = 0.5 Γ 0.4 = 0.2
(ii) π π΅|π΄ =π π΄β©π΅
π(π΄) =
0.2
0.3=
2
3
(πππ) π π΄πΆ βͺ π΅πΆ = π π΄ β© π΅ πΆ
= 1 β π(π΄ β© π΅)
= 1 β 0.2 = 0.8
Example19 Cards are dealt one by one from a well shuffled pack of 52 cards until an ace
appears. Show that the probability that exactly π cards are dealt before the first ace
appears is 4 51βπ 50βπ 49βπ
52.51.50.49
Solution: Let π΄: Drawing π non ace cards
π΅: Drawing an ace in π + 1 π‘β draw
π π΄ = 48πΆπ
52πΆπ
= 48!
π ! 48βπ !.π ! 52βπ !
52!
=(52βπ) 51βπ 50βπ 49βπ
52.51.50.49
Also π π΅|π΄ = 4πΆ1
(52βπ)πΆ1
= 4
52βπ
β΄π π΄ β© π΅ = π(π΄). π π΅|π΄
=4 51βπ 50βπ 49βπ
52.51.50.49
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Example20 Two dice are thrown and sum of numbers appearing is observed to be 6. Find
the conditional probability that number 2 has appeared at least once.
Solution: Let π΄: Number 2 has appeared at least once
π΅: Sum of numbers on two dice is 6
Now π π΄|π΅ = π(π΄β©π΅)
π(π΅)
π΄ β‘ 1,2 , 2,2 , 3,2 , 4,2 , 5,2 , 6,2 , 2,1 , 2,3 , 2,4 , 2,5 , (2,6)
π΅ β‘ 1,5 , 2,4 , 3,3 , 4,2 , (5,1)
π΄ β© π΅ β‘ 2,4 , 4,2
Now π(π΄ β© π΅) = 2
36 and π π΅ =
5
36 β΄π π΄|π΅ =
2
5
Example21 A bag contains 7 red 5 black balls; another bag contains 5 red and 8 black
balls. A ball is drawn from the first bag and without noticing its colour is put in the
second bag and then a ball is drawn from second bag. Find the probability that ball
drawn is red in colour.
Solution: Case 1 π΄: Red ball is drawn from first bag.
π΅: Red ball is drawn from second bag.
π π΄ β© π΅ = π(π΄). π π΅
π΄
=7
12 .
6
14 =
42
168
Case 2 πΆ: Black ball is drawn from first bag.
π΅: Red ball is drawn from second bag.
Then π πΆ β© π΅ = π(πΆ). π π΅
πΆ
=5
12 .
5
14 =
25
168
Required probability = π π΄ β© π΅ + π πΆ β© π΅
= 42
168+
25
168=
67
168
2.5 Bayβs Theorem
If πΈ1,πΈ2,β¦β¦ . ,πΈπ are π mutually exclusive and exhaustive events in a sample space
such that π πΈπ > 0 for each π, (π = 1,2,β¦ , π) and π΄ is an arbitrary event for which
π(π΄) β 0, then the conditional probability of occurrence of πΈπ , given that π΄ has already
occurred is given by
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π πΈπ π΄ = π πΈπ π π΄ πΈπ
π πΈπ π π΄ πΈπ π=ππ=1
, π = 1,2,β¦ ,π
Proof: As πΈ1 ,πΈ2 ,β¦ ,πΈπ are π mutually exclusive and exhaustive events
β΄πΈ1 βͺ πΈ2 βͺ . . .βͺ πΈπ = π
Now π΄ = π΄ β© π = π΄ β© (πΈ1 βͺ πΈ2 βͺ . . .βͺ πΈπ )
βπ΄ = (π΄β©πΈ1) βͺ (π΄β©πΈ2)βͺ(π΄β©πΈ3) βͺ. .. βͺ (π΄β©πΈπ)
By addition law of probability
π(π΄) = π(π΄β©πΈ1) + π(π΄β©πΈ2) +P(π΄β©πΈ3) + β―+P(π΄β©πΈπ)
β΅(π΄β©πΈ1), (π΄β©πΈ2), (π΄β©πΈ3)are all mutually exclusive
βπ π΄ = π πΈ1 π π΄ πΈ1 + π πΈ2 π π΄ πΈ2 + π πΈ3 π π΄ πΈ3 + β―+
π πΈπ π π΄ πΈπ
= π πΈπ π π΄ πΈπ π=ππ=1 β¦β
β΄π πΈπ π΄ = π πΈπβ©π΄
π(π΄) =
π πΈπ π π΄ πΈπ
π πΈπ π π΄ πΈπ π=ππ=1
using β
Example22 Three urns I, II, III contain 6 red and 4 black balls, 2 red and 6 black balls,
and 1 red and 8 black balls respectively. An urn is chosen randomly and a ball is drawn
from the urn. If the ball drawn is red, find the probability that it was drawn from urn I.
Solution: Let π΄: A Red ball is drawn
πΈ1: Urn I is chosen, π(πΈ1) = 1
3, P (π΄ πΈ1) =
6
10 =
3
5
πΈ2: Urn II is chosen, π(πΈ2) = 1
3, P (π΄ πΈ2) =
2
8 =
1
4
πΈ3: Urn III is chosen, π(πΈ3) = 1
3, P (π΄ πΈ3) =
1
9
Required Probability = π(πΈ1 π΄ ) = π πΈ1 π π΄ πΈ1
π πΈ1 π π΄ πΈ1 +π πΈ2 π π΄ πΈ2 +π πΈ3 π π΄ πΈ3
=
1
3 .
3
5
1
3 .
3
5 +
1
3 .
1
4 +
1
3 .
1
9 =
3
5.
180
173=
108
173
Example23 Two urns I and II contain 3 red and 4 black balls, 2 red and 5 black balls
respectively. A ball is transferred from urn I to urn II and then a ball is drawn from urn II.
If the ball drawn is found to be red, find the probability that the ball transferred from urn
I is red.
Solution: Let π΄: A Red ball is drawn from urn II
πΈ1: Ball transferred from urn I is red, π(πΈ1) = 3
7, P (π΄ πΈ1) =
3
8
12 | P a g e
πΈ2: Ball transferred from urn I is black, π(πΈ2) = 4
7, P (π΄ πΈ2) =
2
8
Required Probability = π(πΈ1 π΄ ) = π πΈ1 π π΄ πΈ1
π πΈ1 π π΄ πΈ1 +π πΈ2 π π΄ πΈ2
= 3
7 .
3
8
3
7 .
3
8 +
4
7 .
2
8 =
9
9+8=
9
17
Example24 An insurance company insured 2000 scooter drivers, 4000 car drivers and
6000 truck drivers. The probability of an accident involving a scooter driver is 0.01, a car
driver is 0.03 and a truck driver is 0.15. If one of an insured person meets with an
accident, what is the probability that he is a car driver?
Solution: Let π΄: An insured person meets with an accident
πΈ1: Person is an insured scooter driver
π(πΈ1) = 2000
2000+4000+6000=
2
12, P (π΄ πΈ1) = 0.01
πΈ2: Person is an insured car driver
π(πΈ2) = 4000
2000+4000+6000 =
4
12, P (π΄ πΈ2) = 0.03
πΈ3: Person is an insured truck driver
π(πΈ3) = 6000
2000+4000+6000 =
6
12, P (π΄ πΈ3) = 0.15
Required Probability = π(πΈ2 π΄ ) = π πΈ2 π π΄ πΈ1
π πΈ1 π π΄ πΈ1 +π πΈ2 π π΄ πΈ2 +π πΈ3 π π΄ πΈ3
=4
12 (0.03)
2
12 (0.01)+
4
12 (0.03)+
6
12 (0.15)
=3
26
Example25 Ram speaks truth 2 out of 3 times and Shyam 4 out of 5 times; they agree in
an assertion that from a bag containing 6 balls of different colour, a red ball has been
drawn. Find the probability that the statement is true.
Solution: Let π΄: Ram and Shyam agree in the assertion that a red ball has been drawn
πΈ1: Red ball is drawn. π(πΈ1) = 1
6, P (π΄ πΈ1) =
2
3 .
4
5
πΈ2: Non Red ball is drawn. π(πΈ1) = 5
6, P (π΄ πΈ2) =
1
3 .
1
5
Required Probability= π(πΈ1 π΄ ) =π πΈ1 π π΄ πΈ1
π πΈ1 π π΄ πΈ1 +π πΈ2 π π΄ πΈ2
= 1
6 .
2
3 .
4
5
1
6 .
2
3 .
4
5 +
5
6 .
1
3 .
1
5 =
8
8+5=
8
13
13 | P a g e
Example26 A card from a deck of 52 cards is missing. 2 cards are drawn from the
remaining deck of 51 cards and are found to be of spade. Find the probability that
missing card is of spade.
Solution: Let π΄: 2 cards of spade are drawn from a deck of 51 cards.
πΈ1: Missing card is of spade, π(πΈ1) = 13
52, π(π΄ πΈ1) =
12πΆ2
51πΆ2
=12Γ11
51Γ50
πΈ2: Missing card is a non- spade, π(πΈ2) = 39
52, π(π΄ πΈ2) =
13πΆ2
51πΆ2
=13Γ12
51Γ50
Required Probability = π(πΈ1 π΄ ) = π πΈ1 π π΄ πΈ1
π πΈ1 π π΄ πΈ1 +π πΈ2 π π΄ πΈ2
= 13Γ12Γ11
52Γ51Γ5013Γ12Γ11
52Γ51Γ50+
39Γ13Γ12
52Γ51Γ50
= 11
11+39=
11
50
2.6 Random Variable and Probability Distributions
A random variable is often described as a variable whose values are determined by
chance. The values taken by the variable may be countable or uncountable, based on
which it is classified as discrete or continuous. Random variable is typically denoted
using capital letters such as βπβ.
2.6.1 Discrete Probability Distributions
A discrete probability distribution βπ(π)βdescribes the probability of occurrence of each
value of a discrete random variable βπβ. A discrete random variable is a random variable
that has countable values, such as a set of positive integers. The discrete probability
distribution βπ(π)β of an experiment provides the corresponding probabilities to all
possible values of the random variable βπβ associated with it such that π π = 1
Example27 Find the probability distribution of the number of aces when two cards are
drawn at random with replacement from a well shuffled pack of 52 cards.
Solution: Let π be a random variable showing number of aces. Clearly π can take values
0, 1 or 2. If π denotes success i.e. getting an ace and πΉ denotes failure i.e. getting a non-
ace card, then π π = 4
52=
1
13 and π πΉ =
12
13
π Event π(π)
0 πΉπΉ 12
13 .
12
13=
144
169
1 SF or FS 1
13.12
13 +
12
13.
1
13 =
24
169
2 ππ 1
13 .
1
13=
1
169
14 | P a g e
Example28 Find the probability distribution of the number of successes in two tosses of
a die, when success is defined as a number greater than 4.
Solution: Let π be a random variable showing number of successes in two tosses of a
die. Clearly π can take values 0, 1 or 2 . If π denotes success i.e. getting a number
greater than 4 and πΉ denotes failure, then π π = 2
6=
1
3 and π πΉ =
2
3
π Event π(π)
0 πΉπΉ 2
3 .
2
3=
4
9
1 SF or FS 1
3.2
3 +
2
3.1
3 =
4
9
2 ππ 1
3 .
1
3=
1
9
Example29 3 bad articles are mixed with 7 good ones. Find the probability distribution
of number of bad articles if 3 are drawn at random without replacement from the lot.
Solution: Let π be a random variable showing number of bad articles. Clearly π can
take values 0, 1, 2 or 3.
π Event π(π)
0 0 Bad 3 Good 7πΆ3
10πΆ3
=210
720
1 1 Bad 2 Good 3πΆ1 .7πΆ2
10πΆ3
=378
720
2 2 Bad 1 Good 3πΆ2 .7πΆ1
10πΆ3
=126
720
3 3 Bad 0 Good 3πΆ3
10πΆ3
=6
720
Note: Combination is being used as articles are drawn without replacement.
2.6.2 Mathematical Expectation
If π be a random variable which can assume any one of the values π₯1 , π₯2 ,β―, π₯π with
respective probabilities π1 , π2 , β―, ππ ; then the mathematical expectation of π usually
called as expected value of π, denoted by πΈ(π) is defined as:
πΈ π = π1 π₯1 + π2 π₯2 + β―+ππ π₯π = ππ π₯π ; where ππ = 1
15 | P a g e
Physical interpretation of π¬ πΏ
If π₯ denotes mean of set of observations π₯1 , π₯2 ,β―, π₯π with respective frequencies π1 , π2
, β―, ππ ; then π₯ = ππ π₯π
π , π = ππ
β π₯ = π1
ππ₯1+
π2
ππ₯2 + β―+
ππ
ππ₯π β―β
Now out of total π cases, ππ are favorable to π₯π
β΄ π(π = π₯π ) =ππ
π= ππ , π = 1, 2,β―, π
β π1
π= π1,
π2
π= π2, β―,
ππ
π= ππ β―β‘
β΄ π₯ = π1π₯1 + π2π₯2 + β―+ πππ₯π= ππ π₯π using β‘ in β
Hence π₯ = πΈ(π)
Hence mathematical expectation of a random variable is nothing but its arithmetic mean.
β΄We conclude Mean (π₯ ) = πΈ(π) = ππ π₯π
Similarly Variance ( Ο2 ) = ππ π₯π
2
π β
ππ π₯π
π
2
= ππ π₯π 2 β ππ π₯π
2= πΈ π2 β πΈ(π) 2
Example30 What is the expected number of heads appearing when a fair coin is tossed 3
times.
Solution: Letπ be a random variable showing number of heads. Clearly π can take
values 0, 1, 2 , 3.
π₯π Event ππ ππ π₯π
0 πππ 1
8 0
1 π»ππ,ππ»π,πππ» 3
8
3
8
2 π»π»π,π»ππ»,ππ»π» 3
8
6
8
3 π»π»π» 1
8
3
8
β΄ πΈ(π) = ππ π₯π = 12
8 = 1.5
Example31 A man draws 2 balls from a bag containing 3 white and 5 black balls. If he
receives Rs70 for every white ball he draws and Rs. 35 for every black ball, what is his
expectation?
16 | P a g e
Solution: Following table shows the amount received by the man for each event:
Event Probability (ππ ) Amount (π₯π ) ππ π₯π
2 black balls 5πΆ2
8πΆ2
=10
28
35+35=70 10
28Γ 70
1white 1 black ball 3πΆ1
Γ 5πΆ1
8πΆ2
=15
28
70+35=105 15
28Γ 105
2 white balls 3πΆ2
8πΆ2
=3
28
70+70=140 3
28Γ 140
β΄ πΈ(π) = ππ π₯π = 700
28+
1575
28+
420
28 =
385
4= 96.25
Example32 For a random variable π, the probability mass function is
π π₯ = ππ₯, for π₯ = 1, 2,β―, π
= 0, otherwise
Find expectation of π.
Solution: Here π π₯ denotes probability mass function
β΄ π(π₯)ππ₯=1 = ππ₯π
π₯=1 = π π₯ = 1ππ₯=1
βππ(π+1)
2= 1
βπ = 2
π(π+1)
β΄πΈ π = π₯π(π₯)ππ₯=1 = π₯ππ₯π
π₯=1 = π₯2ππ₯=1
2
π(π+1)
=2
π(π+1) π₯2π
π₯=1
=2
π(π+1) 12 + 22 + β―+ π2
=2
π(π+1)
π π+1 (2π+1)
6 =
(2π+1)
3
Example 33 A random variable π has the following probability function:
π -2 -1 0 1 2 3
π(π) k 0.1 0.3 2k 0.2 K
Calculate mean and variance.
Solution: For any probability function ππ=1
β k + 0.1 + 0.3 + 2k + 0.2 + k = 1
β k = 0.1
17 | P a g e
π₯π ππ πππ₯π πππ₯π2
-2 0.1 -0.2 0.4
-1 0.1 -0.1 0.1
0 0.3 0 0
1 0.2 0.2 0.2
2 0.2 0.4 0.8
3 0.1 0.3 0.9
ππ π₯π = 0.6 ππ π₯π2 = 2.4
β΄Mean = ππ π₯π = 0.6
Variance ( Ο2 ) = ππ π₯π 2 β ππ π₯π
2
= 2.4 β 0.36 = 2.04
2.6.3 Continuous Probability Distributions
The probability distribution π π associated with a continuous random variable π is
called a continuous distribution. A continuous random variable is having a set of infinite
and uncountable values, for example set of real numbers in the interval (0,1) is
uncountable.
If π be a continuous random variable taking values in the interval [π, π], the function
π(π₯) is said to be the probability density function (PDF) of π, if it satisfies the following
properties:
i. π π₯ β₯ 0 βπ₯βπ in [π, π]. ii. Total area under the probability curve is one, i.e. π π β€ π₯ β€ π =1.
For two distinct points π and π in the interval [π, π]; π π β€ π₯ β€ π is givenby area
under the probability curve between the ordinates π₯ = π and π₯ = π, π. π. π π β€ π₯ β€
π = β« ππ
π π₯ ππ₯.
Also πΉ π₯ = π π β€ π₯ = β« ππ₯
ββ π₯ ππ₯ is called cumulative distribution function or
simply distribution function.
Example 34 Letπ be a random variable with PDF given by
π π₯ = ππ₯4 , π₯ β€ 1
= 0, otherwise
i. Find the value of the constant k
ii. Find πΈ π and πππ π
18 | P a g e
iii. Find π π β₯1
2
Solution: π. For π π₯ to be PDF, β« π π₯ ππ₯ = 1β
ββ
ββ« ππ₯4 ππ₯ = 11
β1
β π
5 π₯5 β1
1 = 1 β π = 5
2
ππ. πΈ π = β« π₯π π₯ ππ₯β
ββ
= β« π₯ππ₯4 ππ₯1
β1
= π β« π₯5 ππ₯1
β1= 0 (β΅ π₯5 is an odd function)
Also πππ π = πΈ π2 β πΈ(π) 2
= β« π₯2 π π₯ ππ₯ β 0β
ββ
= π β« π₯6 ππ₯ = 5
7
1
β1
πππ. π π β₯1
2= β« ππ₯4 ππ₯
11
2
= π
5 π₯5 1
2
1 = 31
64
Example 35 Show that the function π(π₯) defined as
π π₯ = πβπ₯ , π₯ β₯ 0
0 , π₯ < 0 , is a probability density function and find the
probability that the variate π having π(π₯) as density function will lie in the
interval (1,2). Also find the probability distribution function πΉ 2 .
Solution: We have β« πβ
ββ π₯ ππ₯ = β« 0
0
ββππ₯ + β« πβπ₯β
0ππ₯ = β πβπ₯ 0
β = 1
β΄π π₯ is a probability density function.
Also π 1 β€ π₯ β€ 2 = β« πβπ₯2
1ππ₯ = β πβπ₯ 1
2 = 2.33
Again πΉ 2 = β« π2
ββ π₯ ππ₯ = β« 0
0
ββππ₯ + β« πβπ₯
2
0ππ₯ = β πβπ₯ 0
2 = 1 β πβ2
= 1 β 0.135 = 0.865
Exercise 2
1. In a single throw of two dice, what is the probability of getting six as a product of
numbers on two dice?
2. A bag contains 4 black, 2 red and 3 blue pens. If 2 pens are drawn at random from
the pack and then another pen is drawn without replacement, what is the
probability of drawing 2 black and 1 blue pens?
19 | P a g e
3. A purse contains 2 silver and 4 copper coins. A second purse contains 4 silver and
3 copper coins. If a coin is pulled out at random from one of the two purses, what
is the probability that coin is a silver coin?
4. A problem in mathematics is given to three students π΄, π΅and πΆ whose chances of
solving are 1
2 ,
1
3 and
1
4 respectively. What is the probability that the problem will
not be solved?
5. If four whole numbers taken at random are multiplied together, find the
probability that the last digit in the product is 1,3,7 or 9.
6. π΄ speaks truth in 60% and π΅ in 75% cases of the cases. In what percentage of
cases they are likely to contradict each other in stating the same fact.
7. π΄ and π΅ take turns in throwing two dice and the first to throw 10 wins the game. If
π΄ has the first throw, find π΅βs chances of winning.
8. π΄ bag contains 10 white and 3 black balls. Another bag contains 3 white and 5
black balls. Two balls are transferred from the first bag and placed in the second
and then one ball is drawn from the second bag. What is the probability that it is a
white ball?
9. In each of a set of games it is 2 to 1in favour the winner of the previous game.
What is the probability that who wins the first game shall win at least 3 games out
of the next four games?
10. π΄ and π΅ take turns in throwing two dice with π΄ having the first trial. The first to
throw 10 being awarded a prize. Find their expectations if the prize money is Rs
460.
11. In a lottery π tickets are drawn out of π tickets numbered from 1 to π. Show that
the expectation of sum of numbers drawn is π π+1
2 .
12. Find the probability distribution of number of kings drawn when two cards are
drawn one by one, without replacement, from a deck of 52 cards.
13. π΄ random variable π has the following probability distribution:
π 0 1 2 3 4 5
π(π) 0.1 πΎ 0.2 2πΎ 0.3 πΎ
Find π.πΎ ππ.π(π β€ 1) πππ.π(π > 3)
Answers
1. 1
9
2. 1
14
3. 19
42
4. 1
4