Let A & B be two events. Suppose `P(A) = 0.4` , `P(B) = p` and `P(AuuB)=0.7` The value of `p` for which `A` and `B` are independent is
A. `1//3`
B. `1//4`
C. `1//2`
D. `1//5`
1 Answers
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Given two events `Aa n dB` . If odds against `A` are as 2:1 and those in favour of `AuuB` are 3:1, then find the range of `P(B)dot`
Clearly P(A) = `1//3`, `P(A uu B) = 3//4`. Now, `P(B) le P(A uu B)` `implies P(B) le 3//4 " "(1)` Also, P(B) = `P(A uu B) - P(A) +...
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Let A,B, C be three mutually independent events. Consider the two statements `S_(1)and S_(2).` `{:(S_(1):A and B nnC "are independent.",),(S_(2):A and
Correct Answer - A We are given that `P(AnnB)=P(A)P(B)` `P(BnnC)=P(B)P(C)` `P(CnnA)=P(C)P(A)` `P(AnnBnnC)=P(A)(B)P(C)` We have, `P(Avv(BnnC)=P(A)(B)P(C)=P(A)P(B)P(C)=P(A)P(BnnC)` `impliesA and BnnC` are independent Therefore, `S_(2)` is true. Also, `P[(Ann(BuuC)]=P[(AnnB)uu(AnnB)nn(AnnC)]` `=P(AnnB)+P(AnnC)-P(AnnBnnC)` `=P(A)P(B)+P(A)P(C)-P(A)P(B)P(C)` `P(A)[P(B)+P(C)-P(B)P(C)]` `P(A)[P(B)+P(C)-P(BnnC)]` `=P(A)P(BnnC)`
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Events `Aa n dC` are independent. If the probabilities relating `A ,B ,a n dC` are `P(A)=1//5, P(B)=1//6; P(AnnC)=1//20 ; P(BuuC)=3//8.` Then events `
Correct Answer - A `P(AnnC)=P(A)P(C)` `or1/20=1/5P(C)` `orP(C)=1/4` Now, `P(BuuC)=1/6+1/4-P(BnnC)` `P(BnnC)=5/12-3/8=1/24P(BnnC)` Therefore, B and C aer independent.
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Let S and T are two events difined on a sample space with probabilities `P(S)=0.5,P(T)=069,P(S//T)=0.5` Events S and T are
Correct Answer - B `P(S//T)=(P(SnnT))/(P(T))` `or 0.5=(P(SnnT))/(0.69)` ` or 0.5=(P(SnnT))/(0.69)` `or P(SnnT)=0.5xx0.69=P(S)P(T)` Therefore, S and T are independent. `thereforeP(Aand T)=P(S)P(T)` ` =0.69xx0.5=0.345` `P(SorT)=P(S)+P(T) -P(SnnT)` `=0.5+0.69-0.345 =08450`
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Suppose A and B are two events with `P(A) = 0.5 and `P(AuuB)=0.8LetP(B)=p` if A and B are mutually exclusive and P(B)=q if A and B are independent eve
Correct Answer - 2 When A and B are mutually exclusive, `P(AnnB)=0` `thereforeP(AuuB)=P(A)+P(B)" "(1)` `implies 0.8 =0.5+5` `or p=0.3" "(2)` `P(AuuB)=P(A)+P(B)=P(A)+P(B)-P(AuuB)=P(A)+P(B)-P(A)P(B)` `or 0.8=0.5+q-(0.5)q` `or 0.3=q//2` `or q//p=2`
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Let A and B be two events such that `P (bar(AuuB))=1/6, P(AnnB)=1/4and P(barA)=1/4,where barA` stands for the complement of the event A. Then the even
Correct Answer - C `P (bar(AuuB))=1/6` `P(AuuB)=5/6,P(A)=3/4` `P(AuuB)-P(A)+P(B)-P(AnnB)=5/6` `impliesP(B)=5/6-3/4+1/4=1/3` `impliesP(AnnB)=P(A).P(B)=1/4`
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Let E and F be two independent events. The probability that exactly one of them occurs is `11//25` and the probability of none of them occurring is `2
Correct Answer - A::D Let `P(E)=eand P(F)=f` `P(EuuF)-P(EnnF)=11/25` `impliese+f-2ef=11/25" "(1)` `P(barEnnbarF)=2/25` `implies(1-e)(1-f)=2/25" "(2)` From (1) and (2), `ef=12/25and e+f=7/5` Solving, we get `e=4/5,f=3/5or e=3/5,f=4/5`
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The probablities of events , `AnnB`, `A`, `B` and `AuuB` are respectively in `A.P.` with second term equal to the common difference. Therefore `A` and
Correct Answer - A::D `(a,d)` `P(AnnB)=a`, `P(A)=a+d`, `P(B)=a+2d`, and `P(AuuB)=a+3d` also `a+d=d` `implies a=0` `implies P(AnnB)=0`, `P(A)=d`, `P(B)=2d` an `P(AuuB)=3d`
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Suppose `Aa n dB` are two events with `P(A)=0. 5a n dP(AuuB)=0. 8.` Let `P(B)=p` if `Aa n dB` are mutually exclusive and `P(B)=q` if `Aa n dB` are ind
Correct Answer - C `(c )` `P(auuB)=P(A)+P(B)-P(AnnB)` ………`(i)` When `A` and `B` are mutually exclusive, then `P(AnnB)=0` `:.0.8=0.5+p` `:.p=0.3` When `A` and `B` are independent `P(AnnB)=P(A)*P(B)` `:.` From `(i)` , `0.8=0.5+q-(0.5)q`...
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For any events `A` and `B`. Given `P(AuuB)=0.6`, `P(A)=P(B)`, `P(B//A)=0.8`. Then the value of `P[AnnbarB)uu(barAnnB)]` is
Correct Answer - D `(d)` Let `P(A)=P(B)=p` (say) `P(B//A)=(P(BnnA))/(P(A))` `:.(0.8)p=P(BnnA)`………..`(i)` `P(AuuB)=p+p-P(AnnB)` `0.6=2p-p(0.8)=(1.2)p` `:.p=(0.6)/(1.2)=(1)/(2)` `:.P(A)=P(B)=(1)/(2)` Now, `P[AnnbarB) uu (barAnnB)]` `=P(A)+P(B)-2P(AnnB)` `=1-2(0.8)(1)/(2)=0.2=(1)/(5)`
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