Since events A, B, C are mutually exclusive and P(A) + P(B) + P(C) =1,
regardless of what type of event X is, we must have
P(A|X) + P(B|X) + P(C|X) = 1. Thus sum condition can be combined with Bayes's theorem to find the desired conditional probabilities.
Bayes's theorem is:
P(A|X) = P(X|A) P(A) / P(X) ;
P(B|X) = P(X|B) P(B) / P(X) ;
P(C|X) = P(X|C) P(C) / P(X)
Plugging in the given probabilities, the sum condition gives
1 = .75 x .4 /P(X) + .5 x .4 /P(X) + .5 x .2 /P(X)
This equation can be solved for P(X) yielding P(X) = .6
Plugging this into the Bayes formulas gives:
P(A|X) = .75 x .4 / .6 = 0.5 (= 1/2)
P(B|X) = .5 x .4 / .6 = .3333 ( = 1/3)
P(C|X) = .5 x .2 / .6 = /16666 (= 1/6)