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CBSEClass–12Mathematics

NCERTsolution

Chapter-13

Probability-Exercise13.2

1.IfP(A)= andP(B)= findP(A B)ifAandBareindependentevents.

Ans.AsAandBareindependentevents.

Therefore, =P(A).P(B)

=

2.Twocardsaredrawnatrandomandwithoutreplacementfromapackof52playing

cards.Findtheprobabilitythatboththecardsareblack.

Ans.S=52cards =52

Twocardsaredrawnwithoutreplacement.

A={26blackcards} =26

P(A)=

AndP(B)i.e.,probabilitythatsecondcardisblackknownthatfirstcardisblack=

P(AandB)=P(A).P(B)=

3.Aboxoforangesisinspectedbyexaminingthreerandomlyselectedorangesdrawn

withoutreplacement.Ifallthethreeorangesaregood,theboxisapprovedforsale

otherwiseitisrejected.Findtheprobabilitythataboxcontaining15orangesoutof

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which12aregoodand3arebadoneswillbeapprovedforsale.

Ans.S={12goodoranges,3badoranges}

=15

Probabilitythatfirstorangedrawnisgood=

Probabilitythatsecondorangeisdrawnisgood=

Probabilitythatthirdorangeisdrawnisgoodwhenboththefirstandsecondaregood=

P(aboxisapproved)=

4.Afaircoinandanunbiaseddiearetossed.LetAbetheevent‘headappearsonthe

coin’andBbetheevent‘3onthedie’.CheckwhetherAandBareindependentevents

ornot.

Ans.S={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

=12

Headappearsonthecoin=A={(H,1),(H,2),(H,3),(H,4),(H,5),(H,6)} =6

P(A)=

3appearsonthedie=B={(H,3),(T,3)} =2

P(B)=

Now(A B)={(H,3)} =1

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AgainP(A).P(B)=

Therefore, P(A).P(B),i.e.,eventsAandBareindependent.

5.Adiemarked1,2,3inredand4,5,6ingreenistossed.LetAbetheevent‘numberis

even’andBbetheevent‘numberisred’.AreAandBindependent?

Ans.S= =6

Thenumberiseven=A={2,4,6} =3

P(A)=

Thenumberisred=B= =3

P(B)=

(A B)={2} =1

NowP(A).P(B)=

Therefore, P(A).P(B),i.e.,AandBarenotindependent.

6.LetEandFbeeventswithP(E)= P(F)= andP(E F)= AreEandF

independent?

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Ans.P(E)= P(F)=

NowP(E).P(F)=

P(E).P(F)

Therefore,EandFarenotindependentevents.

7.GiventhattheeventsAandBaresuchthatP(A)= P(A B)= andP(B)=

Find iftheyare(i)mutuallyexclusive,(ii)independent.

Ans.P(A)= = ,P(B)=

(i)AandBaremutuallyexclusiveevents,then

=P(A)+P(B)

(ii)AandBareindependentevents.

=P(A)+P(B)–P(A).P(B)

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8.LetAandBindependentevents,P(A)=0.3andP(B)=0.4.Find:

(A)P(A B)

(B)P(A B)

(C)P

(D)P

Ans.P(A)=0.3,P(B)=0.4

AandBareindependentevents.

(i) P(A).P(B)=0.3x0.4=0.12

(ii) =P(A)+P(B)–P(A).P(B)=0.03+0.4–0.3x0.4=0.7–0.12=0.58

(iii)P = = =P(A)=0.3

(iv)P = = =P(B)=0.4

9.IfAandBaretwoeventssuchthatP(A)= P(B)= andP(A B)= findP

(notAandnotB).

Ans.P(A)= P(B)=

P(notA)=1–P(A)=

P(notB)=1–P(B)=

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NowP(A).P(B)=

=P(A).P(B)

Thus,AandBareindependentevents.

Therefore,‘notA’and‘notB’areindependentevents.

Hence,P(notAandnotB)=P(notA).P(notB)

=

10.EventsAandBaresuchthatP(A)= P(B)= andP(notAornotB)= State

whetherAandBareindependent.

Ans.

P(A)= andP(B)=

P(A).P(B)=

Therefore, P(A).P(B),i.e.,AandBarenotindependent.

11.GiventwoindependenteventsAandBsuchthatP(A)=0.3,P(B)=0.6.Find:

(A)P(AandB)

(B)P(AandnotB)

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(C)P(AorB)

(D)P(neitherAnorB)

Ans.P(A)=0.3,P(B)=0.6

AandBareindependentevents.

(i)P(AandB)=P(A).P(B)=0.3x0.6=0.18

(ii)P(AandnotB)= =P(A)– =0.3–0.18=0.12

(iii)P(AorB)=P(A)+P(B)–P(AandB)=0.3+0.6–0.18=0.9–0.18=0.72

(iv)P(neitherAnorB)=P[not(A B)]=1–P(A B)=1–0.72=0.28

12.Adieistossedthrice.Findtheprobabilityofgettinganoddnumberatleastonce.

Ans.S={1,2,3,4,5,6} =6

LetArepresentsanoddnumber.

A={1,3,5} =3

P(A)=

=1–P(A)=

=3

NowP(atleastonesuccess)=1–P(anoddnumberonnoneofthethreedice)

=

13.Twoballsaredrawnatrandomwithreplacementfromaboxcontaining10black

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and8redballs.Findtheprobabilitythat:

(i)bothballsarered.

(ii)firstballisblackandsecondisred.

(iii)oneofthemisblackandotherisred.

Ans.S=(10blackballs,8redballs) ==18

Letdrawingofaredballbeasuccess.

A={8redballs} =8

P(A)=

And =1–P(A)=

(i)P(bothareredball)=P(A).P(B)=

(ii)P(firstisblackballandsecondisred)= .P(A)=

(iii)P(oneofthemisblackandotherisred)= .P(A)+P(A).

=

14.ProbabilityofsolvingspecificproblemindependentlybyAandBare and

respectively.Ifbothtrytosolvetheproblemindependently,findtheprobabilitythat:

(i)theproblemissolved.

(ii)exactlyoneofthemsolvestheproblem.

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Ans.P(A)= P(B)= , = and

(i)P(theproblemissolved)=1–P(theproblemisnotsolved)

=

=

=

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

=

15.Onecardisdrawnatrandomfromawellshuffleddeckof52cards.Inwhichofthe

followingcasesaretheeventsEandFindependent?

(i)E:'thecarddrawnisaspade'

F:'thecarddrawnisanace'

(ii)E:'thecarddrawnisblack'

F:'thecarddrawnisaking'

(iii)E:'thecarddrawnisakingorqueen'

F:'thecarddrawnisaqueenorjack'

Ans.S={Allthe52cards} =52

(i)E={13spades} =13

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P(E)=

F={4aces} =4

P(F)=

NowE F={Anaceofspade} =1

=

Also,P(E).P(F)=

Therefore, =P(E).P(F)

Hence,EandFareindependentevents.

(ii)E={26blackcards} =26

P(E)=

F={4kings} =4

P(F)=

NowE F={2blackkings} =2

=

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Also,P(E).P(F)=

Therefore, =P(E).P(F)

Hence,EandFareindependentevents.

(iii)E={4kings,4queens} =8

P(E)=

F={4queens,4jacks} =8

P(F)=

NowE F={4queens} =4

=

Also,P(E).P(F)=

Therefore, P(E).P(F)

Hence,EandFarenotindependentevents.

16.Inahostel60%ofthestudentsreadHindinewspaper,40%readEnglishnewspaper

and20%readbothHindiandEnglishnewspapers.Astudentisselectedatrandom.

(a)FindtheprobabilitythatshereadsneitherHindinorEnglishnewspapers.

(b)IfshereadsHindinewspaper,findtheprobabilitythatshereadsEnglish

newspaper.

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(c)IfshereadsEnglishnewspapers,findtheprobabilityshereadsHindinewspaper.

Ans.LetArepresentsthestudentreadingHindinewspaperandBrepresentsthestudents

readingEnglishnewspaper.

P(A)= P(B)= andP(AandB)=

(a)P(neitherHindinorEnglishnewspaperisread)=1–P(AorB)

=1–[P(A)+P(B)–P(AandB)]

=

=

(b)

P

(c)P

Choosethecorrectanswerinthefollowing:

17.Theprobabilityofobtaininganevenprimenumberoneachdiewhenapairofdice

isrolledis:

(A)0(B) (C) (D)

Ans. =36

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LetArepresentsanevenprimenumberoneachdice.

A={2,2} =1

P(A)=

Henceoption(D)iscorrect.

18.TwoeventsAandBaresaidtobeindependent,if:

(A)AandBaremutuallyexclusive.

(B)P(A’B’)=[1–P(A)][1–P(B)]

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

(D)P(A)+P(B)=1

Ans.P(A’andB’)

=[1–P(A)].[1–P(B)]

=P(A’).P(B’)

Hence,option(B)iscorrect