**Question **: An event X can take place in conjuction with any one of the mutually exclusive and exhaustive events A, B and C. 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 probability of X taking place in conjuction with C is

(A) 5/8

(B) 3/8

(C) 5/24

(D) none of these

**Answer **: option (A)

**Solution **:

P(A)=P(B)=P(C) and P(A)+P(B)+P(C)=1

therefore, P(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)

5/12=1/3{3/8+1/4+P(X/C)}

on solving we get P(X/C)=5/8.