If A, B, C are mutually exclusive and exhaustive events of a sample space S such that P(B)= 3 2 P(A) and P(C)= 1 2 P(B) , then P(A)=

Step-by-step solution

Given mutually exclusive and exhaustive events A , B , and C , the sum P(A) + P(B) + P(C) = 1 holds. The problem provides P(B) = 3 2 P(A) and P(C) = 1 2 P(B) . Substituting the first expression into the second, P(C) = 1 2 ( 3 2 P(A) ) = 3 4 P(A) . Replacing P(B) and P(C) in the sum, P(A) + 3 2 P(A) + 3 4 P(A) = 1 . Combining terms with a common denominator yields 13 4 P(A) = 1 , so P(A) = 4 13 . Among the options, 4 13 corresponds to choice A .