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.