Passage: Let A , B , C and D be mutually exclusive and exhaustive events and P(A) 2 = P(B) 3 = P(C) 5 = P(D) 8 . Question: If G is the geometric mean of P(A) , P(B) , P(C) and P(D) , then what is 9G equal to ?

Step-by-step solution

Since A , B , C and D are mutually exclusive and exhaustive events, the sum of their probabilities is 1 . P(A) + P(B) + P(C) + P(D) = 1 Let P(A) 2 = P(B) 3 = P(C) 5 = P(D) 8 = k . Then P(A) = 2k , P(B) = 3k , P(C) = 5k , P(D) = 8k . Substituting these into the sum equation gives: 2k + 3k + 5k + 8k = 1 18k = 1 k = 1 18 The probabilities are P(A) = 2 18 , P(B) = 3 18 , P(C) = 5 18 , P(D) = 8 18 . The geometric mean G of the four probabilities is: G = (P(A) P(B) P(C) P(D))^ 1 4 G = ( 2 18 3 18 5 18 8 18 )^ 1 4 G = ( 2