An event `X` can take place in conjuction with any one of the mutually exclusive and exhaustive events `A ,Ba n dC` . If `A ,B ,C` are equiprobable and the probability of `X` is 5/12, and the probability of `X` taking place when `A` has happened is 3/8, while it is 1/4 when `B` has taken place, then the probabililty of `X` taking place in conjuction with `C` is a. `5//8` b. `3//8` c. `5//24` d. none of these

Answer

A

Explanation

`P(A) =P(B)=P(C)and P(A)+P(B)+P(C) =1`
`thereforeP(A)=P(B)=P(C)=1/3.`
Also, `P(X)=5/12,P(X//A)=3/8,P(X//B)=1/4`
We have, `P(X)=P(A)P(X//A)+P(B)P(X//B)+P(C) P(X//C)`
`therefore5/12=1/3{(3)/(8)+(1)/(4)+P(X//C)}`
`impliesP(X//C)=5/8`