Solution :
\(P({A’ \cap B’\over C})\) = \(P(A’ \cap B’ \cap C)\over P(C)\)
= \(P(C) – P(A \cap C) – P(B \cap C) + P(A \cap B\cap C)\over P(C)\) ……..(i)
Given, \(P(A \cap B\cap C)\) = 0 and A, B and C are pairwise independent.
\(\therefore\) \(P(A \cap C)\) = P(A).P(C)
and \(P(B \cap C)\) = P(B).P(C)
\(\therefore\) \(P({A’ \cap B’\over C})\) = \(P(C) – P(A).P(C) – P(B).P(C) + 0\over P(C)\)
= 1 – P(A) – P(B) = P(A’) – P(B)
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