(b) \(rac{1}{4}\)Since A and B are mutually exclusive and exhaustive events, therefore, P(A ∩ B) = 0, P(A ∪ B) = 1 We know that, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ⇒ 1 = P(A) + 3 P(A) (Given P(B) = 3P(A)) ⇒ 4P(A) = 1 ⇒ P(A) = \(rac{1}{4}\)∴ P(B) = 1 - \(rac{1}{4}\) = \(rac{3}{4}\) (∵ P(A) + P(B) = 1)⇒ P(\(\bar{B}\)) = 1 - P(B) = 1 - \(rac{3}{4}\) = \(rac{1}{4}\).