NDA Mathematics Question (2026) — Solution

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 = ( 240 18^4 )^ 1 4 Simplifying the expression inside the bracket: G = ( 15 16 16 3^8 )^ 1 4 = ( 15 3^8 )^ 1 4 G = 15^ 1 4 3^2 = 15^ 1 4 9 Multiplying by 9 yields: 9G = 15^ 1 4 Answer: 15^ 1 4