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Let A,B,C be 3 events such that P(A//B)=...

Let `A,B,C` be `3` events such that `P(A//B)=(1)/(5)`, `P(B)=(1)/(2)`, `P(A//C)=(2)/(7)` and `P(C )=(1)/(2)`, then `P(B//A)` is

A

`(4)/(11)`

B

`(5)/(11)`

C

`(6)/(11)`

D

`(7)/(17)`

Text Solution

Verified by Experts

The correct Answer is:
D
`(d)` Assuming the validity of Bayes's theorem.
`P(B//A)=(P(B)*P(A//B))/(P(B)*P(A//B)+P©*P(A//C))`
`=((1)/(2)*(1)/(5))/((1)/(2)*(1)/(5)+(1)/(2)*(2)/(7))=(7)/(17)`
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