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(A) Given that $A, B, C$ are pairwise independent events,we have $P(A \cap B | C) = \frac{P(A \cap B \cap C)}{P(C)}$.Since $P(A \cap B \cap C) = 0$,it

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$P(A') - P(B)$

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(A) Given that $A, B, C$ are pairwise independent events,we have $P(A \cap B | C) = \frac{P(A \cap B \cap C)}{P(C)}$.Since $P(A \cap B \cap C) = 0$,it follows that $P(A \cap B | C) = 0$.We need to find $P(A' \cap B' | C)$.Using De Morgan's Law,$A' \cap B' = (A \cup B)'$.So,$P(A' \cap B' | C) = P((A \cup B)' | C) = 1 - P(A \cup B | C)$.By the Addition Theorem,$P(A \cup B | C) = P(A | C) + P(B | C) - P(A \cap B | C)$.Since $A$ and $C$ are independent,$P(A | C) = P(A)$. Similarly,$P(B | C) = P(B)$.Thus,$P(A \cup B | C) = P(A) + P(B) - 0 = P(A) + P(B)$.Therefore,$P(A' \cap B' | C) = 1 - (P(A) + P(B)) = (1 - P(A)) - P(B) = P(A') - P(B)$.

Let $A, B, C$ be pairwise independent events such that $P(C) > 0$ and $P(A \cap B \cap C) = 0$. Then $P(A' \cap B'|C)$ is equal to:

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