(i) True. A = getting an even number on the first die. B = getting an odd number on the first die. There is no common elements in A and B. ⇒ A ∩ B = ϕ ∴ A and B are mutually exclusive. (ii) True. From (i), A and B are mutually exclusive. A ∪ B = {(1, 1), (1, 2), ...., (1, 6), (2, 1), (2, 2), ..., (2, 6), ..., (6, 1), (6, 2), ..., (6, 6) = S ∴ A ∪ B is mutually exhaustive. (iii) True. B = getting an odd number on the first die. B′ = getting an even number on first die = A. ∴ A = B′ (iv) False. Since A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ ϕ (v) False. Since B′ = A [from (iii)] ∴ A ∩ B′ = A ∩ A = A ≠ ϕ (vi) False. Since A′ = B and B′ = A, A′ ∩ B′ = ϕ B ∩ C = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)} ≠ ϕ A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ ϕ Thus A′, B′ and C are not mutually exclusive.