It is obvious ∵ If A and B are independent, then P(A∩B)=P(A)⋅P(B) By given options, option (a), (b) and (d) are rejected. Method (2) [1−P(A)][1−P(B)]=1−P(A)−P(B)+P(A)⋅P(B) =1−[P(A)+P(B)−P(A)⋅P(B)] =1−[P(A∪B)]{∵P(A∩B)=P(A)⋅P(B)} =P(A∪B) =P(A∩B) =P(AC∩BC)