(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.