Chapter 15 - Probability - SelfStudys

CBSEClass10Mathematics

ImportantQuestions

Chapter 15 -Probability

1. Anintegerischosenatrandomfromthefirsttwohundredsdigit.Whatisthe

probabilitythattheintegerchosenisdivisibleby6or8.

Ans:Multiplesof6first200integers

6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96,102,108,114,120,126,132,138,144,

150,156,162,168,174,180,186,192,198

Multiplesof8first200integers

8,16,24,32,40,48,56,64,72,80,88,96,104,112,120,128,136,144,152,160,168,176,184,

192,200

NumberofMultiplesof6or8=50

P(Multiplesof6or8)=50/200=1/4

2. Aboxcontains12ballsoutofwhichxareblack.ifoneballisdrawnatrandomfrom

theboxwhatistheprobabilitythatitwillbeablackball?If6moreblackballsare

outinthebox.theprobabilityofdrawingablackballisnowdoubleofwhatitwas

before.Findx.

Ans:Randomdrawingofballsensuresequallylikelyoutcomes

Totalnumberofballs=12

Totalnumberofpossibleoutcomes=12

Numberofblackballs=x

('lloutoftotal12outcomes,favourableoutcomes=x

P(blackball)=

(21if6moreblackballsareoutinthebae.thenThetotalnumberofblackballs=x+6

Totalnumberofballsinthebae=12+6=18AccordingtothequestionProbabilityof

drawingblackballissecondcase

=2Xprobabilitydrawingofblackballinfirstcase

6x+36=18x

1

x=3

hencenumberofblackballs=3

3. Abagcontains8redballsandxblueballs,theoddagainstdrawingablueballare2:

5. Whatisthevalueofx?

Ans:No.ofblueballsbex

No.ofredballsbe8Totalno.ofballs=x+8

Probabilityofdrawingblueballs=

Probabilityofdrawingredballs

2x=40.

∴x=20

4. Acardisdrawnfromawellshuffleddeckofcards

i. Whataretheoddsinfavourofgettingspade?

ii. Whataretheoddsagainstgettingaspade?

iii. Whataretheoddsinfavourofgettingafacecard?

iv. Whataretheoddsinfavourofgettingaredking

Ans:Totalcards52

Spade=13

Remainingcards39

i. Theoddsinfavourofgettingspade13

Theoddsisnotinfavourofgettingspade39

ii. Theoddsagainstgettingaspade13

Theoddsnotagainstgettingaspade39

iii. Theoddsinfavourofgettingafacecard12

Theoddsnotinfavourofgettingafacecard40

iv. Theoddsinfavourofgettingaredking2

Theoddsnotinfavourofgettingaredking50

2

5. Adieisthrownrepeatedlyuntilasixcomesup.Whatisthesamplespaceforthis

experiment?HINT:A={6}B={1,2,3,4,5,}

Ans:Thesamplespaceis={A,BA,BBA,BBBA,BBBBA……….})

6. Whyistossingacoinconsideredtobeafairwavofdecidingwhichteamshouldget

theballatthebeginningofafootballmatch?

Ans:equallylikelybecausetheyaremutuallyexclusiveevents.

7. Abagcontains5redballsandsomeblueballs.Iftheprobabilityofdrawingablue

ballisdoublethatofaredball.determinethenumberofblueballsinthebag.

Ans:Letthenumberofblueballsisthebagbex

Thentotalnumberofballsisthebag=5+x

∴Numberofallpossibleoutcomes=5+x

Numberofoutcomesfavourabletotheeventofdrawingablueball=x

(Qtherearexblueballs)

∴Probabilityofdrawingablueball

Similarly,probabilityofdrawingaredball

Accordingtotheanswer

x=10

8. Aboxcontains12ballsoutofwhichxareblack.Ifoneballisdrawnatrandomfrom

thebox.whatistheprobabilitythatitwillbeablackball?If6moreblackballsare

outintheboxtheprobabilityofdrawingablackballisnowdoubleofwhatitwas

before.Findx?

Ans:Numberofallpossibleoutcomes=12

Numberofoutcomesfavourabletotheeventofdrawingblackball=x

Requiredprobability

Nowwhen6moreblackballsareoutinthebox.

Numberofallpossibleoutcomes=12+6=18

Numberofoutcomesfavourabletotheeventofdrawingablackball=x+6

∴Probabilityofdrawingablackball

Accordingtothequestion.

3

∴x=3

9. If65%ofthepopulationshaveblackeves.25%havebrowneyesandtheremaining

haveblueeyes.Whatistheprobabilitythatapersonselectedatrandomhas(i)Blue

eves(ii)Brownorblackeves(iii)Blueorblackeves(iv)neitherbluenorbrowneves

Ans:No.ofblackeves=65

No.ofBrowneves=25

No.ofblueeves=10

Totalno.ofeves=180

i) P(Blueeves)=

ii) P("Brownorblackeves)=

iii) P(Blueorblackeves)=

iv) P(neitherbluenorbrowneves)

10. Findtheprobabilityofhaving53Sundaysin

i. aleapyear

ii. anonleapyear

Ans:Anordinaryyearhas365daysi.e.52weeksand1day

Thisdaycanbeanyoneofthe7daysoftheweek.

∴PithatthisdayisSunday

Hence.P(anordinaryyearhas53Sunday)

Aleapyear366daysi.e.52weeksand2days

Thisdaycanbeanyoneofthe7daysoftheweek

∴P(thatthisdayisSunday)

Hence.P(aleapyearhas53Sunday)

11. Fivecards-theten.Jack,queen,kingandace,arewellshuffledwiththeirface

downwards.Onecardisthennickedunatrandom.

i. Whatistheprobabilitythatthecardisaqueen?

ii. Ifthequeenisdrawnandoutaside,whatistheprobabilitythatthesecondcard

nickedupisa(a)anace(b)aqueen

Ans:Here,thetotalnumberofelementaryevents=5

4

i. Since,thereisonlyonequeen

∴Favourablenumberofelementaryevents=1

∴Probabilityofsettingthecardofqueen

ii. Now.thetotalnumberofelementaryevents=4

(a) Since,thereisonlyoneace


∴Probabilityofgettinganacecard

(b) Since,thereisnoqueen(asqueenisnutaside)


∴Probabilityofgettingaqueen

12. Anumberxischosenatrandomfromthenumbers-3,-2,-1,01,2,3.Whatisthe

probabilitythat|x|<2

Ans:xcantake7values

Toset|x|<2take-1.0.1

Probability(|x|<2)

13. Anumberxisselectedfromthenumbers1,2,3andthenasecondnumbervis

randomlyselectedfromthenumbers1,4,9,Whatistheprobabilitythattheproduct

xvofthetwonumberswillbelessthan9?

Ans:NumberXcanbeselectedinthreewavsandcorrespondingtoeachsuchwavthere

arethreewavsofselectingnumbery.Therefore.twonumberscanbeselectedin9wavs

aslistedbelow:

(l.1),(l.4),(2.l),(2.4),(3.1)


Hence,requiredprobability

14. Intheadjoiningfigureadartisthrownatthedartboardandlandsintheinteriorof

thecircle.Whatistheprobabilitythatthedartwilllandintheshadedregion.

5

Ans:Wehave

AB=CD=8andAD=BC=6

usingPythagorasTheoremisAABC.wehave

AC2=AB2+BC2

AC2=82+62=100

AC=10

OA=OC=5[QOisthemidpointofAC]

∴Areaofthecircle squnits[QArea= r2]

ABCD=AB×BC=8×6=48squnits

Areaofshadedregion=Areaofthecircle-AreaofrectangleABCD

Areaofshadedreeion=25π-48squnit.

Hence

P(Dartlandsintheshadedregion)=

15. InthefigpointsA.B.CandDarethecentresoffourcircles.eachhavingaradiusof

1unit.IfapointischosenatrandomfromtheinteriorofasquareABCD.whatisthe

probabilitythatthepointwillbechosenfromtheshadedregion.

Ans:Radiusofthecircleis1unit

Areaofthecircle=Areaof4sector

SideofthesquareABCD=2units

Areaofsquare=2×2=4units

Areashadedregionis

=Areaofsquare-4×Areaofsectors

Probability=

16. IntheadjoiningfigureABCDisasquarewithsidesoflength6unitspointsP&Oare

themidpointsofthesidesBC&CDrespectively.Ifapointisselectedatrandom

fromtheinteriorofthesquarewhatistheprobabilitythatthepointwillbechosen

fromtheinteriorofthetriangleAPO.

6

Ans:AreaoftrianglePOC =4.5units2

AreaoftriangleABP

AreaoftriangleADQ

AreaoftriangleAPO=Areaofasquare-(AreaofatrianglePQC+AreaoftriangleABP+

AreaoftriangleABP)

=36-(l8+4.5)

=36-22.5

=13.5

ProbabilitythatthepointwillbechosenfromtheinteriorofthetriangleAPO

17. Inamusicalchairsamethepersonplayingthemusichasbeenadvisedtostop

playingthemusicatanytimewithin2minutesaftershestartsplaying.Whatisthe

probabilitythatthemusicwillstopwithinthehalfminuteafterstarting.

Ans:Herethepossibleoutcomesareallthenumbersbetween0and2.

Thisistheportionofthenumberlinefrom0to2asshowninfigure.

LetAbetheeventthat‘themusicisstoppedwithinthefirsthalfminute.’Then,outcomes

favorabletoeventAareallpointsonthenumberlinefromOtoOi.e.

ThetotalnumberofoutcomesarethepointsonthenumberlinefromOtoPi.e.0to2.

∴P(A)=

18. Ajarcontains54marbleseachofwhichisblue.greenorwhite.Theprobabilityof

selectingabluemarbleatrandomfromthejaris andtheprobabilityofselecting

agreenmarbleatrandomis .Howmanywhitemarblesdoesthejarcontain?

7

Ans:Lettherebebblue,ggreenandwwhitemarblesinthemarblesinthejar.Then,b+

g+w=54

∴P(Selectingabluemarble)

Itisgiventhattheprobabilityofselectingabluemarbleis .

Wehave.

P(Selectingagreenmarble)

[QP(Selectingagreenmarble)= (Given)

⇒ g=24

Substitutingthevaluesofbandgin(i),weget

18+24+w=54⇒w=12

8


ImportantQuestions

Chapter15

Probability

1MarksQuestions

1. Completethestatements:

(i) ProbabilityofeventE+Probabilityofevent“notE”=_______________

(ii) Theprobabilityofaneventthatcannothappenis_______________.Suchaneventis

called_______________.

(iii) Theprobabilityofaneventthatiscertaintohappenis_______________.Suchanevent

iscalled_______________.

(iv) Thesumoftheprobabilitiesofalltheelementaryeventsofanexperimentis

_______________.

(v) Theprobabilityofaneventisgreaterthanorequalto_______________andlessthanor

equalto_______________.

Ans.(i)1

(ii) 0,impossibleevent.

(iii) 1,sureorcertainevent

(iv) 1

(v) 0,1

2. Whichofthefollowingcannotbetheprobabilityofanevent:

(A)

(B)

9

(C) 15%

(D) 0.7

Ans.(B)SincetheprobabilityofaneventEisanumberP(E)suchthat

0 P(E) 1

cannotbetheprobabilityofanevent.

3. IfP(E)=0.05,whatistheprobabilityof‘notE’?

Ans.SinceP(E)+P(notE)=1

P(notE)=1–P(E)=1–0.05=0.95

4. Itisgiventhatinagroupof3students,theprobabilityof2studentsnothavingthe

samebirthdayis0.992.Whatistheprobabilitythatthe2studentshavethesame

birthday?

Ans.LetEbetheeventofhavingthesamebirthday

P(E)=0.992

ButP(E)+P =1

P =1–P(E)=1–0.992=0.008

5. 12defectivepensareaccidentlymixedwith132goodones.Itisnotpossibletojust

lookatapenandtellwhetherornotitisdefective.Onepenistakenoutatrandom

fromthislot.Determinetheprobabilitythatthepentakenoutisagoodone.

Ans.Totalnumberoffavourableoutcomes=132+12=144

Numberoffavourableoutcomes=132

Hence,P(gettingagoodpen)=

10

6. Whichofthefollowingispolynomial?

(a)

(b)

(c)

(d) noneofthese

Ans.(d)noneofthese

7. Polynomial isa

(a) linearpolynomial

(b) quadraticpolynomial

(c) cubicpolynomial

(d) bi-quadraticpolynomial

Ans.(d)bi-quadraticpolynomial

8. If and arezerosof ,thenthevalueof is

(a) 5

(b) -5

(c) 8

(d) -8

Ans.(b)-5

9. Thesumandproductofthezerosofaquadraticpolynomialare2and-15

11

respectively.Thequadraticpolynomialis

(a)

(b)

(c)

(d)

Ans.(b)

10. Cardseachmarkedwithoneofthenumbers4,5,6,…20areplacedinaboxandmixed

thoroughly.Onecardisdrawnatrandomfromthebox,whatistheprobabilityof

gettinganevenprimenumber?

(a) 0

(b) 1

(c) 2

(d) 3

Ans.(a)0

11. Abagcontains5redand4blackballs.Aballisdrawnatrandomfromthebag.What

istheprobabilityofgettingablackball?

(a)

(b)

(c)

(d) Noneofthese

Ans.(c)

12

12. Adiceisthrownonce,whatistheprobabilityofgettingaprimenumber?

(a) 1

(b)

(c)

(d) 0

Ans.(b)

13. Whatistheprobabilitythatanumberselectedfromthenumbers1,2,3,…15isa

multipleof4?

(a)

(b)

(c)

(d) 1

Ans.(a)

14. Cardsmarkedwiththenumbers2to51areplacedinaboxandmixedthroughly.

Onecardisdrawnfromthisbox,findtheprobabilitythatthenumberonthecardisan

evennumber.

(a)

(b) 1

(c)

(d) Noneofthese

13

Ans.(a)

15. Theking,queenandjackofclubsareremovedfromadeckof52playingcardsand

thenwellshuffled.Onecardisselectedfromtheremainingcards,findtheprobability

ofgettingaking.

(a)

(b) 1

(c)

(d) noneofthese

Ans.(a)

16. Whatistheprobabilityofgettinganumberlessthan7inasinglethrowofadie?

(a)

(b) 0

(c) 1

(d) noneofthese

Ans.(c)1

17. Onecardisdrawnfromawellshuffleddeckof52cards.Findtheprobabilityof

drawing‘10’ofablacksuit.

(a)

(b) 1

(c)

(d) 0

14

Ans.(a)

18. Cardseachmarkedwithoneofthenumbers4,5,6,…20areplacedinaboxandmixed

thoroughly.Onecardisdrawnatrandomfromthebox,whatistheprobabilityof

gettinganevenprimenumber?

(a) 0

(b) 1

(c) 2

(d) 3

Ans.(a)0

19. Abagcontains5redand4blackballs.Aballisdrawnatrandomfromthebag.What

istheprobabilityofgettingablackball?

(a)

(b)

(c)

(d) Noneofthese

Ans.(c)

20. Adiceisthrownonce,whatistheprobabilityofgettingaprimenumber?

(a) 1

(b)

15

(c)

(d) 0

Ans.(b)

21. Whatistheprobabilitythatanumberselectedfromthenumbers1,2,3,…15isa

multipleof4?

(a)

(b)

(c)

(d) 1

Ans.(a)

22. IfEbeanyevent,thenvalueofP(E)lieinbetween

(a)

(b)

(c)

(d)

Ans.(c)

23. Maximumandminimumvalueofprobabilityis

(a) (1,1)

(b) (1,0)

(c) (0,1)

16

(d) noneofthese

Ans.(b)(1,0)

24. Anunbiaseddieisthrown.Whatistheprobabilityofgettinganevennumberora

multipleof3?

(a)

(b)

(c) 1

(d) noneofthese

Ans.(a)

25. LetEbeanyevent,thenthevalueofP(E)+P(notE)equalsto

(a) 1

(b) 0

(c) 3

(d)

Ans.(a)1

26. Degreeofpolynomialy3–2y2- is

(a)

(b) 2

(c) 3

(d)

Ans.(c)3

17

27. ZerosofP(x)=2x2+9x–35are

(a) 7and

(b) -7and

(c) 7and5

(d) 7and2

Ans.(b)-7and

28. Thequadraticpolynomialwhorezerosare3and-5is

(a) x2+2x–15

(b) x2+3x–8

(c) x2–5x–15

(d) Noneofthese

Ans.(a)x2+2x–15

29. If and arethezerosofthequadraticpolynomialP(x)=x2–px+q,thenthe

valueof isequalto

(a) p2–2q

(b)

(c) q2–2p

(d) noneofthese

Ans.(a)p2–2q

18


ImportantQuestions

Chapter15

Probability

2MarksQuestions

1. Whichofthefollowingexperimentshaveequallylikelyoutcomes?Explain.

(i) Adriverattemptstostartacar.Thecarstartsordoesnotstart.

(ii) Aplayerattemptstoshootabasketball.She/heshootsormissestheshot.

(iii) Atrialismadetoansweratrue-falsequestion.Theanswerisrightorwrong.

(iv) Ababyisborn.Itisaboyoragirl.

Ans.(i)Intheexperiment,“Adriverattemptstostartacar.Thecarstartsordoesnotstart”,

wearenotjustifiedtoassumethateachoutcomeisaslikelytooccurastheother.Thus,the

experimenthasnoequallylikelyoutcomes.

(ii) Intheexperiment,“Aplayerattemptstoshootabasketball.She/heshootsormissesthe

shot”,we are not justified to assume that each outcome is as likely to occur as the other.

Thus,theexperimenthasnoequallylikelyoutcomes.

(iii) Intheexperiment“Atrialismadetoansweratrue-falsequestion.Theanswerisrightor

wrong.”We know, in advance, that the result can lead in one of the two possibleways –

eitherrightorwrong.Wecanreasonablyassumethateachoutcome,rightorwrong,islikely

tooccurastheother.

Thus,theoutcomesrightorwrongareequallylikely.

(iv) Intheexperiment,“Ababyisborn,Itisaboyoragirl”.Weknow,inadvancethatthe

outcomecanleadinoneofthetwopossibleoutcomes–eitheraboyoragirl.Wearejustified

toassumethateachoutcome,boyorgirl,islikelytooccurastheother.Thus,theoutcomes

boyorgirlareequallylikely.

19

2. Whyistossingacoinconsideredtobeafairwayofdecidingwhichteamshouldget

theballatthebeginningofafootballgame?

Ans.Thetossingofacoinisconsideredtobeafairwayofdecidingwhichteamshouldget

theballatthebeginningofafootballgameasweknowthatthetossingofthecoinonlyland

inoneof twopossibleways–eitherheadupor tailup. Itcanreasonablybeassumedthat

eachoutcome,headortail,isaslikelytooccurastheother,i.e.,theoutcomesheadandtail

areequallylikely.Sotheresultofthetossingofacoiniscompletelyunpredictable.

3. A bag contains lemon flavoured candles only.Malini takes out one candywithout

lookingintothebag.Whatistheprobabilitythatshetakesout:

(i) anorangeflavouredcandy?

(ii) alemonflavouredcandy?

Ans.(i)Consider theeventrelated to theexperimentof takingoutofanorange flavoured

candyfromabagcontainingonlylemonflavouredcandies.

Sincenooutcomegivesanorangeflavouredcandy,therefore,itisanimpossibleeventsoits

probabilityis0.

(ii) Considertheeventoftakingalemonflavouredcandyoutofabagcontainingonlylemon

flavouredcandies.Thiseventisacertaineventsoitsprobabilityis1.

4. Abagcontains3redballsand5blackballs.Aballisdrawnatrandomfromthebag.

Whatistheprobabilitythattheballdrawnis:

(i) red?

(ii) notred?

Ans.Thereare3+5=8ballsinabag.Outofthese8balls,onecanbechosenin8ways.

Totalnumberofelementaryevents=8

(i) Sincethebagcontains3redballs,therefore,oneredballcanbedrawnin3ways.

20

Favourablenumberofelementaryevents=3

HenceP(gettingaredball)=

(ii) Sincethebagcontains5blackballsalongwith3redballs,thereforeoneblack(notred)

ballcanbedrawnin5ways.


HenceP(gettingablackball)=

5. Aboxcontains5redmarbles,8whitemarblesand4greenmarbles.Onemarbleis

takenoutoftheboxatrandom.Whatistheprobabilitythatthemarbletakenoutwill

be:

(i) red?

(ii) white?

(iii) notgreen?

Ans.Totalnumberofmarblesinthebox=5+8+4=17


(i) Thereare5redmarblesinthebox.


P(gettingaredmarble)=

(ii) Thereare8whitemarblesinthebox.


P(gettingawhitemarble)=

21

(iii) Thereare5+8=13marblesinthebox,whicharenotgreen.


P(notgettingagreenmarble)=

6. Apiggybankcontainshundred50pcoins,fiftyRe.1coins,twentyRs.2coinsandten

Rs.5coins.Ifitisequallylikelythatofthecoinswillfalloutwhenthebankisturned

upsidedown,whatistheprobabilitythatthecoin:

(i) willbea50pcoin?

(ii) willnotbeaRs.5coin?

Ans.Totalnumberofcoinsinapiggybank=100+50+20+10=180


(i) Thereareonehundred50coinsinthepiggybank.


P(fallingoutofa50pcoin)= =

(ii) Thereare100+50+20=170coinsotherthanRs.5coin.


P(fallingoutofacoinotherthanRs.5coin)= =

7. Gopibuysafishfromashopforhisaquarium.Theshopkeepertakesoutonefishat

randomfromatankcontaining5malefishesand8femalefishes(seefigure).Whatis

theprobabilitythatthefishtakenoutisamalefish?

22

Ans.Totalnumberoffishinthetank=5+8=13


Thereare5malefishesinthetank.


Hence,P(takingoutamalefish)=

8. Fivecards–thenten,jack,queen,kingandaceofdiamonds,arewell-shuffledwith

theirfacedownwards.Onecardisthenpickedupatrandom.

(i) Whatistheprobabilitythatthecardisthequeen?

(ii) If thequeen isdrawnandputaside,what is theprobability that the secondcard

pickedupis(a)anace?(b)aqueen?

Ans.Totalnumberoffavourableoutcomes=5

(i) Thereisonlyonequeen.

Favourableoutcome=1

Hence,P(thequeen)=

(ii) Inthissituation,totalnumberoffavourableoutcomes=4

(a) Favourableoutcome=1

23

Hence,P(anace)=

(b) Thereisnocardasqueen.

Favourableoutcome=0

Hence,P(thequeen)=

9. (i)Alotof20bulbscontains4defectiveones.Onebulbisdrawnatrandomfromthe

lot.Whatistheprobabilitythatthisbulbisdefective?

(ii) Supposethebulbdrawnin(i)isnotdefectiveandisnotreplaced.Nowonebulbis

drawnatrandomfromtherest.Whatistheprobabilitythatthisbulbisnotdefective?

Ans.(i)Totalnumberoffavourableoutcomes=20


HenceP(gettingadefectivebulb)=

(ii) Nowtotalnumberoffavourableoutcomes=20–1=19

Numberoffavouroableoutcomes=19–4=15

HenceP(gettinganon-defectivebulb)=

10. Aboxcontains90discswhicharenumbered from1 to90. Ifonedisc isdrawnat

randomfromthebox,findtheprobabilitythatitbears(i)atwo-digitnumber

(ii) aperfectsquarenumber

(iii) anumberdivisibleby5.


24

(i) Numberoftwo-digitnumbersfrom1to90are90–9=81

Favourableoutcomes=81

Hence,P(gettingadiscbearingatwo-digitnumber)=

(ii) From1to90,theperfectsquaresare1,4,9,16,25,36,49,64and81.


HenceP(gettingaperfectsquare)=

(iii) Thenumbersdivisibleby5from1to90are18.


HenceP(gettinganumberdivisibleby5)=

11. Achildhasadiewhosesixfacesshowthelettersasgivenbelow:

A,B,C,D,E,A

Thedieisthrownonce.Whatistheprobabilityofgetting:

(i) A?

(ii) D?


(i) Numberoffavourableoutcomes=2

HenceP(gettingaletterA)=

(ii) Numberoffavourableoutcomes=1

25

HenceP(gettingaletterD)=

12. Supposeyoudropadieat randomon the rectangular region shown in the figure

givenonthenextpage.What is theprobabilitythat itwill landinsidethecirclewith

diameter1m?

Ans.Totalareaofthegivenfigure(rectangle)=3x2=6m2

AndAreaofcircle= = = m2

Hence,P(dietolandinsidethecircle)=

13. Alotconsistsof144ballpensofwhich20aredefectiveandtheothersaregood.Nuri

willbuyapenifitisgood,butwillnotbuyifitisdefective.Theshopkeeperdrawsone

penatrandomandgivesittoher.Whatistheprobabilitythat:

(i) shewillbuyit?

(ii) shewillnotbuyit?


(i) Numberofnon-defectivepens=144–20=124


HenceP(shewillbuy)=P(anon-defectivepen)=

26

(ii) Numberoffavourableoutcomes=20

HenceP(shewillnotbuy)=P(adefectivepen)=


ballisdoublethatofaredball,determinethenumberofblueballsinthebag.

Ans.Lettherebe blueballsinthebag.

Totalnumberofballsinthebag=

Now,P1=Probabilityofdrawingablueball=

AndP2=Probabilityofdrawingablueball=

Butaccordingtoquestion,P1=2P2

=2x

=2

Hence,thereare10blueballsinthebag.

15. Aboxcontains12ballsoutofwhich areblack. If oneball isdrawnat random

fromthebox,whatistheprobabilitythatitwillbeablackball?

If6moreblackballsareputinthebox,theprobabilityofdrawingablackballisnow

doubleofwhatitwasbefore.Find

Ans.Thereare12ballsinthebox.

27

Therefore,totalnumberoffavourbaleoutcomes=12

Thenumberoffavourableoutcomes=

Therefore,P1=P(gettingablackball)=

If6moreballsputinthebox,then

Totalnumberoffavourableoutcomes=12+6=18

AndNumberoffavourableoutcomes=

P2=P(gettingablackball)=

Accordingtoquestion,P2=2P1

=2x

=2

16. Ajarcontains24marbles,somearegreenandothersareblue.Ifamarbleisdrawn

atrandomfromthejar,theprobabilitythatitisgreenis Findthenumberifblue

ballsinthejar.

Ans.Here,Totalnumberoffavourableoutcomes=24

Lettherebe greenmarbles.

Therefore,Favourablenumberofoutcomes=

P(G)=

28

ButP(G)=

=16

Therefore,numberofgreenmarblesare16

Andnumberofbluemarbles=24–16=8

17. Whyistossingacoinconsideredisthewayofdecidingwhichteamshouldgetthe

ballatthebeginningofafootballmatch?

Ans. Probabilityofhead

Probabilityoftail

i.e.

Probabilityofgettingheadandtailbotharesame.

Tossingacoinconsideredtobefairway.

18. Anunbiaseddieisthrown,whatistheprobabilityofgettinganevennumber?

Ans.Totalnumberofoutcomesare1,2,3,4,5and6,whichare6innumberfavourablecase=

1

Requiredprobability=

19. Twounbiasedcoinsaretossedsimultaneously,findtheprobabilityofgettingtwo

29

heads.

Ans.TotalnumberofoutcomesareHH,HT,TH,TT,whichare4innumbers

Favourableoutcomes=HH,whichisonly


20. Onecardisdrawnfromawellshuffleddeckof52cards.Findtheprobabilityof

gettingajackofhearts.

Ans.Totalnumberofoutcomes=52

Favorablecases=1[Thereisonlyonejackofhearts]


21. Agameconsistsoftossingaone-rupeecoin3timesandnotingitsoutcomeeach

time.Hanifwinsifallthetossesgivethesameresulti.e.,threeheadsorthreefailsand

losesotherwise.CalculatetheprobabilitythatHanifwilllosethegame.

Ans.Sinceacoinistossed3times

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

3headsand3tails{HHH,TTT}

(3headsand3tails)=2

P(Hanifwillwinthegame)=

P(Hanifwilllosethegame)=

30

22. Gopybuysafishfromashopforhisaquarium.Theshopkeepertakeoutonefishat

randomfromatankcontaining5malefishand8femalefish.Whatistheprobability

thatthefishtakenoutisamalefish?

Ans.Totalno.offishes=5+8=13

No.ofmalefishes=5

P(malefish)=

23. Alotconsistsof144ballpensofwhich20aredefectiveandtheothersarefood.Arti

willbuyapenifitisgoodbutwillnotbuyifitisdefective.Theshopkeeperdrawsone

penatrandomandgivesittoher.Whatistheprobabilitythat

(i) shewillbuyit

(ii) shewillnotbuyit?

Ans.Totalno.ofballpens=144

Numberofdefectivepens=20

Numberofgoodpens=144–20=124

(i) P(shewillbuyit)=

(ii) P(shewillnotbuy)=

24. Harpreettossestwodifferentcoinssimultaneously(sayoneisofRs1andotherisRs

2),whatistheprobabilitythatshegets“atleastonehead”?

Ans.Fortossing2coins

Totalpossibleoutcomesare{HH,TT,TH,HT}=4

31

Favourableoutcomes=atleastonehead={TH,HT,HH}=3


25. Whyistossingacoinconsideredisthewayofdecidingwhichteamshouldgetthe

ballatthebeginningofafootballmatch?

Ans. Probabilityofhead

Probabilityoftail

i.e.

Probabilityofgettingheadandtailbotharesame.

Tossingacoinconsideredtobefairway.

26. Twounbiasedcoinsaretossedsimultaneously,findtheprobabilityofgettingtwo

heads.

Ans.TotalnumberofoutcomesareHH,HT,TH,TT,whichare4innumbers

Favourableoutcomes=HH,whichisonly


27. One card is drawn from awell shuffled deck of 52 cards. Find the probability of

gettingajackofhearts.


32

Favorablecases=1[Thereisonlyonejackofhearts]


28. Iftwodicearethrownonce,findtheprobabilityofgetting9.

Ans.Totalnumberofpossibleoutcomesofthrowingtwodice=

Numberofoutcomesofgetting9i.e.,(3+6),(4+5),(5+4),(6+3)=4

Requiredprobability

29. Acardisdrawnfromawellshuffleddeckofplayingcards.Findtheprobabilityof

gettingafacecard.

Ans.Totalnumberofpossibleoutcomes=52

Favourableoutcomes=4+4+4=12[4jack,4queen,4king]


30. Whatistheprobabilityofhaving53Mondaysinaleapyear?

Ans.Totalnumberofdaysinaleapyear=366

Thiscontains52weeksand2days

TheremainingtwodaysmaybeMT,TW,WTh,ThF,FS,SS,SM

FavourablecasesareMT,SMi.e.,2outof7cases


31. Cardsbearingnumbers3to20areplacedinabagandmixedthoroughly.Acardis

33

takenoutfromthebagatrandom,whatistheprobabilitythatthenumberonthecard

takenoutisanevennumber?

Ans.Totalnumberofoutcomes=20–3=17

Cardsintheboxhavingevennumbersare4,6,8,10,12,14,16,18,20,whichare9innumber

favourableoutcomes=9

P(anevennumber)=

34


ImportantQuestions

Chapter15

Probability

3MarksQuestions

1. Agameofchanceconsistsofspinninganarrowwhichcomestorestpointingatone

of thenumbers1, 2, 3, 4, 5, 6, 7, 8 (see figure)and theseareequally likelyoutcomes.

Whatistheprobabilitythatitwillpointat:

(i) 8?

(ii) anoddnumber?

(iii) anumbergreaterthan2?

(iv) anumberlessthan9?

Ans.Outof8numbers,anarrowcanpointanyofthenumbersin8ways.

Totalnumberoffavourableoutcomes=8

(i) Favourablenumberofoutcomes=1

Hence,P(arrowpointsat8)=

(ii) Favourablenumberofoutcomes=4

35

Hence,P(arrowpointsatanoddnumber)=

(iii) Favourablenumberofoutcomes=6

Hence,P(arrowpointsatanumber>2)=

(iv) Favourablenumberofoutcomes=8

Hence,P(arrowpointsatanumber<9)= =1

2. Adiceisthrownonce.Findtheprobabilityofgetting:

(i) aprimenumber.

(ii) anumberlyingbetween2and6.

(iii) anoddnumber.

Ans.Totalnumberoffavourableoutcomesofthrowingadice=6

(i) Onadice,theprimenumbersare2,3and5.

Therefore,favourableoutcomes=3

HenceP(gettingaprimenumber)=

(ii) Onadice,thenumberlyingbetween2and6are3,4,5.


HenceP(gettinganumberlyingbetween2and6)=

(iii) Onadice,theoddnumbersare1,3and5.


36

HenceP(gettinganoddnumber)=

3. Agameconsistsoftossingaonerupeecoin3timesandnotingitsoutcomeeachtime.

Hanifwinsifallthetossesgivethesameresult,i.e.,threeheadsorthreetailsandloses

otherwise.CalculatetheprobabilitythatHanifwilllosethegame.

Ans.Theoutcomesassociatedwiththeexperimentinwhichacoinistossedthrice:

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

Therefore,Totalnumberoffavourableoutcomes=8


Hencerequiredprobability=

4. Adieisnumberedinsuchawaythatitsfacesshowthenumbers1,2,2,3,3,6.Itis

thrown two timesand the total score in two throws isnoted.Complete the following

tablewhichgivesafewvaluesofthetotalscoreonthetwothrows:

Whatistheprobabilitythatthetotalscoreis:

(i) even

(ii) 6

(iii) atleast6?

37

Ans.Completetableisasunder:

Itisclearthattotalnumberoffavourableoutcomes=6x6=36

(i) Numberoffavourableoutcomesofgettingtotalscoreevenare18

HenceP(gettingtotalscoreeven)=

(ii) Numberoffavourableoutcomesofgettingtotalscore6are4

HenceP(gettingtotalscore6)=

(iii) Numberoffavourableoutcomesofgettingtotalscoreatleast6are15

HenceP(gettingtotalscoreatleast6)=

5. 18 cards numbered 1, 2, 3,…18 are put in a box andmixed thoroughly. A card is

drawnatrandomfromthebox.Findtheprobabilitiesthatthecardbears

(i) anevennumber

(ii) anumberdivisibleby2or3

Ans.Totalno.ofpossibleoutcomes=18

(i) Favorablecasesare2,4,6,8,10,12,14,16,18i.e.,9innumber


38

(ii) Favorablecaresare2,3,4,6,8,9,10,12,14,15,16,18i.e.,12innumber


6. Abagcontains5redballs,4greenballsand7whiteballs.Aballisdrawnatrandom

fromthebox.Findtheprobabilitythattheballdrownis

(a) white

(b) neitherrednorwhite

Ans.Totalnumberofballsinthebag=5+4+7=16

Totalnumberofpossibleoutcomes=16

(a) Favourableoutcomesforawhiteball=7


(b) Favourableoutcomesforneitherrednorwhiteball=Numberofgreenballs=4


7. Aboxcontains20ballsbearingnumbers1,2,3,4,…20.Aballisdrawnatrandomfrom

thebox,whatistheprobabilitythatthenumberontheballis

(i) anoddnumber

(ii) divisibleby2or3

(iii) primenumber


(i) Favorableoutcomesare1,3,5,7,9,11,13,15,17,19i.e.,10innumber.

39


(ii) Number“divisibleby2”are2,4,6,8,10,12,14,16,18,20i.e.,10innumber

Numbers“divisibleby3are3,6,9,12,15,18.i.e.,6innumber

Numbers“divisibleby2or3are6,12,18i.e.,3innumber.

Numbersdivisibleby“2or3”=10+6–3=13



(iii) Primenumbersare2,3,5,7,11,13,17,19i.e.,8innumber



8. Abagcontains5redandsomeblueballs,

(i) ifprobabilityofdrawingablueballfromthebagistwicethatofaredball,findthe

numberofblueballsinthebag.

(ii) ifprobabilityofdrawingablueballfromthebagisfourtimesthatofaredball,find

thenumberofblueballsinthebag.

Ans.Letnumberofblueballs=

Totalnumberofballs=

Probabilityofredball=

Probabilityofblueball=

40

Bygivencondition,

(i)

No.ofblueballs=10

(ii) Here,

Hence,numberofblueballs=20

9. Aboxcontains3bluemarbles,2whitemarbles.Ifamarbleistakenoutatrandom

fromthebox,whatistheprobabilitythatitwillbeawhiteone?Blueone?Redone?

Ans.Totalno.ofpossibleoutcomes=3+2+4=9

No.offavourableoutcomesforwhitemarbles=2


No.offavourableoutcomesforbluemarbles=3


No.offavourableoutcomesforredmarbles=4


10. Theintegersfrom1to30inclusivearewrittenoncards(onenumberononecard).

These cardoneput inaboxandwellmixed. Josephpickedupone card.What is the

probabilitythathiscardhas

41

(i) number7

(ii) anevennumber

(iii) aprimenumber


(i) P(theno.7)=

(ii) Evenno.are2,4,6,8,10,12,14,16,18,20,22,24,26,28,30



(iii) Primenumbersfrom1to30are2,3,5,7,11,13,17,19,23,29}

No.offavourableoutcomes=10


11. A bag contains lemon flavored candies only.Malini takes out one candywithout

lookingintothebag.Whatistheprobabilitythatshetakesout

(i) anorangeflavoredcandy?

(ii) alemonflavoredcandy?

Ans. Thebaghaslemonflavoredcandiesonly.

(i) P(anorangeflavoredcandy)=

(ii) P(alemonflavoredcandy)=

42


ballfromthebagistwicethatofared,findthenumberofblueballsinthebag.

Ans.Supposeno.ofbluebolls=

Totalno.ofballs=

Probabilityofblueballs=

Probabilityofredballs=

Accordingto,question,

Hence,no.ofblueballs=12

13. A bag contains 5 red, 4 black and 3 green balls. A ball is taken out of the bag at

random,findtheprobabilitythattheselectedballis

(i) ofredcolour,

(ii) notofgreencolour.

Ans.Totalnumberofballsinthebag=5+4+3

Totalnumberofoutcomes=12

(i) No.ofredballs=5

Requiredprobabilityforaredball=

(ii) favourablecasesfornongreenball=12–3=9

43

Requiredprobabilityforanongreenball=

14. Fromawellshuffledpackof52cards,blackacesandblackqueensareremoved.

Fromtheremainingcardsacardisdrawnatrandom,findtheprobabilityofdrawinga

kingoraqueen.

Ans.Totalnumberofcards=52

Numberofblackaces=2

Numberofblackqueens=2

Cardsleft=52–2–2=48

Totalnumberofequallylikelycases=48

Numberofkingsandqueensleftinthe48cards=4+2=6

Favourablecases=6

Requiredprobability= =

15. Whichofthefollowingexperimentshaveequallylikelyoutcomes?Explain.

(i) Adriverattemptstostartacar.Thecarstartsordoesnotstart.

(ii) Aplayerattemptstoshootabasketball,she/heshootsormissestheshot.

(iii) Ababyisborn.Itisaboyoragirl.

Ans.(i)Adriverattemptstostartacar,thencarstartsordoesnotstartarenotequallylikely.

(ii) Aplayerattemptstoshootabasketball,she/heshootsormissestheshootarenotequally

likely.

(iii) Ababyisborn,itisaboyoragirlisanequallylikelyevent.

44

16. Findtheprobabilitythatanumberselectedatrandomfromthenumbers1,2,3,…35

isa

(i) primenumber,

(ii) multipleof7,

(iii) multipleof3or5.

Ans.(i)Primenumberare2,3,5,7,11,13,17,19,23,29,31,whichare11innumber


P(Primenumber)=

(ii) Multiplesof‘7’are7,14,21,28,35,whichare5innumber

P(amultipleof7)=

(iii) Multipleof‘3’are3,6,9,…33,whichare11innumbers

Multipleof5are5,10,15,…35,whichare7innumber

Multipleof‘3’and‘5’are15,30,whichare2innumber

Multipleof3or5=11+7–2=16

P(Multipleof3or5)=

45


ImportantQuestions

Chapter15

Probability

4MarksQuestions

1. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of

getting:

(i) akingofredcolour

(ii) afacecard

(iii) aredfacecard

(iv) thejackofhearts

(v) aspade

(vi) thequeenofdiamonds.


(i) Therearetwosuitsofredcards,i.e.,diamondandheart.Eachsuitcontainsoneking.


Hence,P(akingofredcolour)=

(ii) Thereare12facecardsinapack.


Hence,P(afacecard)=

(iii) Therearetwosuitsofredcards,i.e.,diamondandheart.Eachsuitcontains3facecards.

Favourableoutcomes=2x3=6

46

Hence,P(aredfacecard)=

(iv) Thereareonlyonejackofheart.

Favourableoutcome=1

Hence,P(thejackofhearts)=

(v) Thereare13cardsofspade.


Hence,P(aspade)=

(vi) Thereisonlyonequeenofdiamonds.

Favourableoutcome=1

Hence,P(thequeenofdiamonds)=

2. Adieisthrowntwice.Whatistheprobabilitythat:

(i) 5willnotcomeupeithertime?

(ii) 5willcomeupatleastonce?

Ans.(i)Theoutcomesassociatedwiththeexperimentinwhichadiceisthrownistwice:

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

47

Therefore,Totalnumberoffavourableoutcomes=36

Nowconsiderthefollowingevents:

A=firstthroeshows5andB=secondthrowshows5

Therefore,thenumberoffavourableoutcomes=6ineachcase.

P(A)= andP(B)=

P andP


(ii) Let S be the sample space associated with the random experiment of throwing a die

twice.Then,n(S)=36

A B=firstandsecondthrowshoe5,i.e.getting5ineachthrow.

Wehave,A=(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

AndB=(1,5)(2,5)(3,5)(4,5)(5,5)(6,5)

P(A)= ,P(B)= andP(A B)=

Requiredprobability=Probabilitythatatleastoneofthetwothrowsshows5

=P(A B)=P(A)+P(B)–P(A B)

=

3. Two customers Shyam and Ekta are visiting a particular shop in the same week

(TuesdaytoSaturday).Eachisequallylikelytovisittheshoponanydayasonanother

day. What is the probability that both will visit the shop on (i) the same day? (ii)

consecutivedays?(iii)differentdays?

Ans.Totalfavourableoutcomesassociatedtotherandomexperimentofvisitingaparticular

shopinthesameweek(TuesdaytoSaturday)bytwocustomersShyamandExtaare:

48

(T,T)(T,W)(T,TH)(T,F)(T,S)

(W,T)(W,W)(W,TH)(W,F)(W,S)

(TH,T)(TH,W)(TH,TH)(TH,F)(TH,S)

(F,T)(F,W)(F,TH)(F,F)(F,S)

(S,T)(S,W)(S,TH)(S,F)(S<S)

Totalnumberoffavourableoutcomes=25

(i) Thefavourableoutcomesofvisitingonthesamedayare(T,T),(W,W),(TH,TH),(F,F)and

(S,S).



(ii) Thefavourableoutcomesofvisitingonconsecutivedaysare(T,W),(W,T),(W,TH),(TH,

W),(TH,F),(F,TH),(S,F)and(F,S).



(iii) Numberoffavourableoutcomesofvisitingondifferentdaysare25–5=20



4. Acardisdrawnatrandomfromawellshuffleddeckofplayingcards.Findthe

probabilitythatthecarddrawnis

(i) acardofspadesofanace

(ii) aredking

49

(iii) neitherakingnoraqueen

(iv) eitherakingoraqueen

(v) afacecard

(vi) cardswhichisneitherkingnoraredcard.

Ans.Totalpossibleoutcomes=52

(i) No.ofspades=13

No.oface=4

1cardiscommon[aceofspade]

Favourableoutcomes=13+4–1=16


(ii) No.ofredkings=2



(iii) No.ofkingandqueen=4+4=8

Favourableoutcomes=52–8=44


(iv) No.ofkingandqueen=4+4=8


(v) No.offacecards=4+4+4=12[Jack,queenandkingarefacecard]

50


(vi) No.ofcardswhichareneitherredcardnorking=52–(26+4–2

=52–28=24


5. Cardsmarkedwithnumbers1,2,3,…25areplacedinaboxandmixedthoroughlyand

onecardisdrawnatrandomfromthebox,whatistheprobabilitythatthenumberon

thecardis

(i) aprimenumber?

(ii) amultipleof3or5?

(iii) anoddnumber?

(iv) neitherdivisibleby5norby10?

(v) perfectsquare?

(vi) atwo-digitnumber?


(i) favourablecasesare2,3,5,7,11,13,17,19,23whichare9innumber


(ii)Multipleof3or5

Favourablecasesare3,5,6,9,10,12,15,18,20,21,24,25,whichare12innumber


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(iii) Favourablecasesare1,3,5,7,9,11,13,15,17,19,21,23,25,whichare13innumber


(iv) Favourablecasesare1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19,21,22,23,24,whichare20in

number


(v) Perfectsquarenumbersare1,4,9,16,25

Favourablecasesare=5


(vi) Two-digitnumbersare10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25=16


6. Fromapackof52playingcards,jacks,queens,kingsandacesofredcolourare

removed.Fromtheremainingacardisdrawnatrandom.Findtheprobabilitythatthe

carddrawnis(i)ablackqueen,(ii)aredcard,(iii)ablackjack,(iv)ahonorablecard.


Cardsremoved=2+2+2+2=8[2jack,2queen,2kingand2acesofredcolour]

Remainingnumberofcards=52–8=44


(i) Favourableoutcomes=2[Thereare2blackqueen]


52

(ii) Favourableoutcomes=numberofredcardsleft=26–8=18

Probabilityforaredcard=

(iii) Favourableoutcomes=Numberofblackjacks=2


(iv) Numberofpicturecardsleft=2+2+2=6[jack,queen,Kingarepicturecards]


(v) Honorablecards[ace,jack,queenandking]

No.ofhonorablecardsleft=2+2+2+2=8


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Chapter 15 - Probability - SelfStudys

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