Let A, B and C are pairwise independent events with P(C) > 0 and $$P(A \cap B\cap C)$$ = 0. Then $$P(A’ \cap B’/C)$$ is equal to

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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