Let A and B be two events such that P(A∪B)=P(A∩B) and P(A∪B)=P(A)+P(B)−P(A∩B) option (a): since P(A∪B)=P(A∩B) (given) therefore A and B are equally likely Suppose option (b) and option (c) are correct. ∴P(A∩B′)=0 and P(A′∩B)=0 ⇒P(A)−P(A∩B)=0 and P(B)−P(A∩B)=0 ⇒P(A)=P(A∩B) and P(B)=P(A∩B) Thus P(A)=P(B)=P(A∩B)=P(A∪B)[∵ Given P(A∩B)=P(A∪B)] Also, we know P(A∪B)=P(A)+P(B)−P(A∩B) =P(A∩B)+P(A∩B)−P(A∩B) =P(A∩B) which is true from given condition Hence, option (a), (b) and (c) are correct.