11th And 12th > Mathematics
PROBABILITY II MCQs
Total Questions : 30
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Answer: Option A. ->
181
:
A
∑8x=0P(X)=1⇒P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)=1=λ+3λ+5λ+7λ+9λ+11λ+13λ+15λ+17λ=1⇒92(λ+17λ)=1⇒81λ=1∴λ=181
:
A
∑8x=0P(X)=1⇒P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)=1=λ+3λ+5λ+7λ+9λ+11λ+13λ+15λ+17λ=1⇒92(λ+17λ)=1⇒81λ=1∴λ=181
Answer: Option D. ->
38
:
D
Let E denote the event that a six occurs and A be the event theat the man reports that it is a six. We have P(E)=16,P(E′)=56,P(AE)=34 and P(AE′)=14. By Baye’s theorem
P(EA)=P(E).P(AE)P(E).P(AE)+P(E′).P(AE′)=(16)(34)(16)(34)+(56)(14)=38
:
D
Let E denote the event that a six occurs and A be the event theat the man reports that it is a six. We have P(E)=16,P(E′)=56,P(AE)=34 and P(AE′)=14. By Baye’s theorem
P(EA)=P(E).P(AE)P(E).P(AE)+P(E′).P(AE′)=(16)(34)(16)(34)+(56)(14)=38
Answer: Option D. ->
(35)7 - (815)7
:
D
Each coupon can be selected in 15 ways. The total number of ways of choosing 7 copouns is 157. If largest number is 9, then the selected numbers have to be from 1 to 9 excluding those consisting of only 1 to 8.
Probability desired is 97−87157
=(35)7−(815)7
:
D
Each coupon can be selected in 15 ways. The total number of ways of choosing 7 copouns is 157. If largest number is 9, then the selected numbers have to be from 1 to 9 excluding those consisting of only 1 to 8.
Probability desired is 97−87157
=(35)7−(815)7
Answer: Option B. ->
The probability that A wins and loses equal number of matches is 1781
:
B
Probability of equal number of W and L is the probability of following
(0)W,(0)L +(1)W,(1)L+(2)W, (2)L
=(13)5+5C1.4C1(13)5+5C2.3C2(13)5=1781
:
B
Probability of equal number of W and L is the probability of following
(0)W,(0)L +(1)W,(1)L+(2)W, (2)L
=(13)5+5C1.4C1(13)5+5C2.3C2(13)5=1781
Answer: Option B. ->
is false
:
B
Since the events A,B,C are mutually exclusive, we have
P(A⋃B⋃C)=23+14+16=1312>0
Which is not possible , hence the statement is false.
:
B
Since the events A,B,C are mutually exclusive, we have
P(A⋃B⋃C)=23+14+16=1312>0
Which is not possible , hence the statement is false.
Answer: Option D. ->
0.37
:
D
Given: P(A)=0.25
P(B)=0.5
P(A and B simultaneously happening) = 0.12
P(A∪B)=P(A)+P(B)−P(A∩B)P(A∪B)=0.25+0.5−0.12=0.63
P(¯A∩¯B)=P¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(A∪B)=1−P(A∪B)=1−0.63=0.37
:
D
Given: P(A)=0.25
P(B)=0.5
P(A and B simultaneously happening) = 0.12
P(A∪B)=P(A)+P(B)−P(A∩B)P(A∪B)=0.25+0.5−0.12=0.63
P(¯A∩¯B)=P¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯(A∪B)=1−P(A∪B)=1−0.63=0.37
Answer: Option A. ->
12
:
A
1+4p4,1−p2,1−2p2 are the probabilities of three mutually exclusive events.
∴0≤1+4pp≤1,0≤1−p2≤1
0≤1−2p2≤1and0≤(1+4p4)+(1−p2)+1−2p2≤1
∴−14≤p≤341−≤p≤1,12≤p≤12andp≤52
⇒p=12
:
A
1+4p4,1−p2,1−2p2 are the probabilities of three mutually exclusive events.
∴0≤1+4pp≤1,0≤1−p2≤1
0≤1−2p2≤1and0≤(1+4p4)+(1−p2)+1−2p2≤1
∴−14≤p≤341−≤p≤1,12≤p≤12andp≤52
⇒p=12
Answer: Option B. ->
14
:
B and C
Probability of r successes in n trials is equal to nCrprqn−r, where p is the probability of success in one trial and q is the probability of failure in one trial.
This means nC4(12)n,nC5(12)n,nC6(12)n are A.P.
⇒2×nC5=nC4+nC6
⇒2×n!(n−5)!5!=n!(n−4)!4!+n!(n−6)!!
⇒6×5+(n−4)(n−5)=2×6(n−4)
⇒n2−9n+50=2(6n−24)
⇒n2−21n+98=0
⇒n=7 or n=14.
:
B and C
Probability of r successes in n trials is equal to nCrprqn−r, where p is the probability of success in one trial and q is the probability of failure in one trial.
This means nC4(12)n,nC5(12)n,nC6(12)n are A.P.
⇒2×nC5=nC4+nC6
⇒2×n!(n−5)!5!=n!(n−4)!4!+n!(n−6)!!
⇒6×5+(n−4)(n−5)=2×6(n−4)
⇒n2−9n+50=2(6n−24)
⇒n2−21n+98=0
⇒n=7 or n=14.
Answer: Option B. ->
37
:
B
Let A be the event that father has at least one boy and B be the event that he has 2boys and one girl.
P(A)=P(1boy,2girl)+P(2boy,1girl)+P(3boy,nogirl)=3C1+3C2+3C323=78
P(A⋂B)=P(2 boy,1 girl)=38
Hence P(BA)=P(A⋂B)P(A)=37.
:
B
Let A be the event that father has at least one boy and B be the event that he has 2boys and one girl.
P(A)=P(1boy,2girl)+P(2boy,1girl)+P(3boy,nogirl)=3C1+3C2+3C323=78
P(A⋂B)=P(2 boy,1 girl)=38
Hence P(BA)=P(A⋂B)P(A)=37.
Answer: Option D. ->
51101
:
D
Let P1 and P2 be the respectively be the probability of heads showing 50 coins and head showing 51 coins P1=100C50 p50(1−p)50 and P2=100C51 p51(1−p)49
Given P1=P2
⇒100C50 p50(1−p)50=100C51 p51(1−p)49
⇒1−pp=5051⇒p=51101.
:
D
Let P1 and P2 be the respectively be the probability of heads showing 50 coins and head showing 51 coins P1=100C50 p50(1−p)50 and P2=100C51 p51(1−p)49
Given P1=P2
⇒100C50 p50(1−p)50=100C51 p51(1−p)49
⇒1−pp=5051⇒p=51101.