Let A and B denote the events that the man wins contracts X and Y respectively. Then P(A) = 0.5 P(A') = 1 - 0.5 = 0.5 P(B') = 0.3 P(B) = 1 - 0.3 = 0.7 (a) The probability that the man wins both contracts = \(0.5 \times 0.7 = 0.35\). (b) The probability that the man wins exactly one of the contracts is \(P(A) \times P(B') + P(B) \times P(A')\) = \(0.5 \times 0.3 + 0.7 \times 0.5\) = \(0.15 + 0.35\) = \(0.50\) (c) Neither of the contracts (i.e not X, not Y) = \(P(A') \times P(B')\) = \(0.5 \times 0.3\) = \(0.15\)