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1.

Probability that A speaks truthis `4/5`. A coin is tossed. A reports that aappears. The probability that actually there was head isA. `4/5`B. `1/2`C. `1/5`D. `2/5`

Answer» Correct Answer - A
Let `E_(1)=` event of getting head,
`E_(2)=` event of not getting head,
and `E=` event that A tells, a head is obtained
`:.P(E_(1))=P(E_(2))=1/2``P(E//E_(1))=` probability that A tells a head is obtained when a head appears `=4//5`
`P(E//E_(2))=` probability that a tells, a head is obtained when a head does not apear
`=1-4/5=1/5`
Now, required probability `=`
`P(E_(1)//E)=(P(E_(1)).P(E//E_(1)))/(P(E_(1)).P(E//E_(1))+P(E_(2)).P(E//E_(2))0`
`=(1/2xx4/5)/(1/2xx4/5+1/2xx1/5)=4/5`
Discussion
2.

If A and B are two events such that `AsubB` and `P(B)!=0` when of the followng is correct?A. `P(A//B)=(P(B))/(P(A))`B. `P(A//B)ltP(A)`C. `P(A//B)geP(A)`D. None of these

Answer» Correct Answer - C
(c)
If `AsubB,` then `AnnB=A`
`:.P(AnnB)=P(A)`
We know that `P(A//B)=(P(AnnB))/(P(B))=(P(A))/(P(B))`
but `P(B)le1:.P(A/B)geP(A)`
Discussion
3.

If A and B are two events such that `P(A)!=0`and `P(B | A) = 1`; then(A) `AsubB` (B) `BsubA` (C) `B=varphi` (D) `A=varphi`A. `AsubB` but `A!=B`B. `BsubA`C. `B=phi`D. `A=phi`

Answer» Correct Answer - A
(a)
Given `P(A)!=0`
and `P(B/A)=1impliesP(B/A)=(P(AnnB))/(P(A))=1`
`impliesP(AnnB)=P(A)impliesAnnB=AimpliesAsubB`
Discussion
4.

In each of the Exercises choose the correct answer:If `P(A) =1/2`, `P (B) = 0`, then `P (A|B)`isA. `0`B. `1/2`C. not definedD. 1

Answer» Correct Answer - C
(c)
Given that
`P(A)=1/2` and `P(B)=0`
`P(A//B)=(P(AnnB))/(P(B))=(P(AnnB))/0=oo`
Therefore `P(A/B)` is not defined.
Discussion