Let A, B, C be pairwise independent events wi | vedclass

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(A) We need to find $P(A' \cap B' | C)$. By the definition of conditional probability:$P(A' \cap B' | C) = \frac{P(A' \cap B' \cap C)}{P(C)}$Using the

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

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(A) We need to find $P(A' \cap B' | C)$. By the definition of conditional probability:$P(A' \cap B' | C) = \frac{P(A' \cap B' \cap C)}{P(C)}$Using the property $P(X \cap Y \cap Z) = P(Z) - P(X' \cap Z) - P(Y' \cap Z) + P(X' \cap Y' \cap Z)$ is not direct,so we use the inclusion-exclusion principle for sets:$P(A' \cap B' \cap C) = P(C \setminus ((A \cup B) \cap C)) = P(C) - P((A \cup B) \cap C)$$P(A' \cap B' \cap C) = P(C) - [P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)]$Given $P(A \cap B \cap C) = 0$ and $A, B, C$ are pairwise independent,we have $P(A \cap C) = P(A)P(C)$ and $P(B \cap C) = P(B)P(C)$.Substituting these values:$P(A' \cap B' \cap C) = P(C) - P(A)P(C) - P(B)P(C) + 0$$P(A' \cap B' \cap C) = P(C)(1 - P(A) - P(B))$Therefore,$P(A' \cap B' | C) = \frac{P(C)(1 - P(A) - P(B))}{P(C)} = 1 - P(A) - P(B)$Since $1 - P(A) = P(A')$,we get $P(A' \cap B' | C) = P(A') - P(B)$.

List-$I$	List-$II$
$A$. $P(E_2)$	$(i)$ $\frac{1}{2}$
$B$. $P(\frac{E_1}{E_2})$	$(ii)$ $\frac{5}{6}$
$C$. $P(\frac{\bar{E}_2}{E_1})$	$(iii)$ $\frac{1}{3}$
$D$. $P(\bar{E}_1 \cup \bar{E}_2)$	$(iv)$ $\frac{1}{6}$
	$(v)$ $\frac{2}{3}$

Let $A, B, C$ be pairwise independent events with $P(C) > 0$ and $P(A \cap B \cap C) = 0$. Then $P(A' \cap B' | C) = $

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