Let 0

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(D) Given $P(A \cup B) = P(A) + P(B) - P(A)P(B).$ By the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B).$ Comparing these,we

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$A$ and $C$ both

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(D) Given $P(A \cup B) = P(A) + P(B) - P(A)P(B).$ By the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B).$ Comparing these,we get $P(A \cap B) = P(A)P(B).$ This implies that events $A$ and $B$ are independent. For independent events $A$ and $B$,the conditional probability is $P(A/B) = P(A).$ Thus,option $(A)$ is correct. Also,if $A$ and $B$ are independent,then $A^c$ and $B^c$ are also independent. By De Morgan's Law,$(A \cup B)^c = A^c \cap B^c.$ Therefore,$P((A \cup B)^c) = P(A^c \cap B^c) = P(A^c)P(B^c).$ Thus,option $(C)$ is also correct. Hence,both $(A)$ and $(C)$ are correct.

Let $0 < P(A) < 1$,$0 < P(B) < 1$ and $P(A \cup B) = P(A) + P(B) - P(A)P(B).$ Then

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