Let A, B and C be three events,which are pair | vedclass

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(C) Given that $A, B$ and $C$ are pairwise independent. Since $A, B, C$ are pairwise independent,$P(A \cap B) = P(A)P(B)$,$P(B \cap C) = P(B)P(C)$,and

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$P(\overline{A}) - P(B)$

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(C) Given that $A, B$ and $C$ are pairwise independent. Since $A, B, C$ are pairwise independent,$P(A \cap B) = P(A)P(B)$,$P(B \cap C) = P(B)P(C)$,and $P(A \cap C) = P(A)P(C)$. Given $P(A \cap B \cap C) = 0$. Since $A, B, C$ are pairwise independent,we have $P(A \cap B \cap C) = P(A \cap B)P(C) = P(A)P(B)P(C) = 0$. Since $P(C) > 0$,it implies $P(A)P(B) = 0$. Now,$P((\overline{A} \cap \overline{B}) / C) = \frac{P(\overline{A} \cap \overline{B} \cap C)}{P(C)}$. Since $A, B, C$ are pairwise independent,the events $\overline{A}, \overline{B}, C$ are also independent. Thus,$P(\overline{A} \cap \overline{B} \cap C) = P(\overline{A})P(\overline{B})P(C)$. Therefore,$P((\overline{A} \cap \overline{B}) / C) = \frac{P(\overline{A})P(\overline{B})P(C)}{P(C)} = P(\overline{A})P(\overline{B})$. $P(\overline{A})P(\overline{B}) = (1 - P(A))(1 - P(B)) = 1 - P(A) - P(B) + P(A)P(B)$. Since $P(A)P(B) = 0$,we get $1 - P(A) - P(B) = P(\overline{A}) - P(B)$.

Let $A, B$ and $C$ be three events,which are pairwise independent and $\overline{E}$ denote the complement of an event $E$. If $P(A \cap B \cap C) = 0$ and $P(C) > 0$,then $P((\overline{A} \cap \overline{B}) / C)$ is equal to

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