We know that, P(A∩B)=P(A)+P(B)−P(A∪B)
Also, P(A∪B)≤1 P(A∩B)min , when P(A∪B)max=1 ⇒P(A∩B)≥P(A)+P(B)−1 ∴ Option (a) is true.
Again P(A∪B)≥0 ∴P(A∩B)max, when P(A∪B)min=0 ⇒P(A∩B)≤P(A)+P(B) ∴ Option (b) is true
Also, P(A∩B)=P(A)+P(B)−P(A∪B.) Thus, (c) is
also correct.
Hence, (a), (b), (c) are correct options .