P(A) = 4/6 = 2/3, P(B) = 3/6 = 1/2 P(AB) = 1/6 P(A).P(B) = 1/3 As P(AB) ≠ P(A).P(B) ∴ A and B are not independent. Occurrence of one do not effect the occurrence of other, P(A ∩ B) ≠ null set ∴ Events are not mutually exclusive. All the possible outcomes are existing, i.e., A ∪ B is equal to sample space {1, 2, 3, 4, 5, 6}, therefore events A and B are exhaustive.