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(C) Given that $A, B, C$ are pairwise independent events,we have $P(A \cap B) = P(A)P(B)$,$P(B \cap C) = P(B)P(C)$,and $P(C \cap A) = P(C)P(A)$.We are

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$\frac{3}{2}-b-c$

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(C) Given that $A, B, C$ are pairwise independent events,we have $P(A \cap B) = P(A)P(B)$,$P(B \cap C) = P(B)P(C)$,and $P(C \cap A) = P(C)P(A)$.We are given $P(\bar{B} \cup \bar{C}) = \frac{1}{2}$.By De Morgan's Law,$P(\bar{B} \cup \bar{C}) = 1 - P(B \cap C) = \frac{1}{2}$,which implies $P(B \cap C) = \frac{1}{2}$.Since $B$ and $C$ are independent,$P(B)P(C) = bc = \frac{1}{2}$.We need to find $P((\bar{B} \cap \bar{C}) \mid A) = \frac{P((\bar{B} \cap \bar{C}) \cap A)}{P(A)}$.Using the property of sets,$(\bar{B} \cap \bar{C}) \cap A = A \setminus (A \cap (B \cup C)) = A \setminus ((A \cap B) \cup (A \cap C))$.Thus,$P((\bar{B} \cap \bar{C}) \cap A) = P(A) - P((A \cap B) \cup (A \cap C)) = P(A) - [P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)]$.Since the events are pairwise independent,$P(A \cap B \cap C)$ is not necessarily $P(A)P(B)P(C)$ unless they are mutually independent. However,the question implies pairwise independence. Assuming mutual independence for the intersection term:$P((\bar{B} \cap \bar{C}) \cap A) = P(A) - P(A)P(B) - P(A)P(C) + P(A)P(B)P(C) = P(A)(1 - b - c + bc)$.Substituting $bc = \frac{1}{2}$,we get $P((\bar{B} \cap \bar{C}) \cap A) = P(A)(1 - b - c + \frac{1}{2}) = P(A)(\frac{3}{2} - b - c)$.Therefore,$P((\bar{B} \cap \bar{C}) \mid A) = \frac{3}{2} - b - c$.

Let $A, B, C$ be three pairwise independent events of a random experiment. If $P(\bar{B} \cup \bar{C}) = \frac{1}{2}$,$P(A) > 0$,$P(B) = b$,and $P(C) = c$,then $P((\bar{B} \cap \bar{C}) \mid A) = $

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