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If $A$ and $B$ are two events such that $P(B) \neq 0$ and $P(B) \neq 1$, then $P(\bar{A} \mid \bar{B})$ is
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The correct answer is:
$\frac{1-P(A \cup B)}{P(\bar{B})}$
$P(\bar{A} \mid \bar{B})=\frac{P(\bar{B} \cap \bar{A})}{P(B)}=\frac{1-P(A \cup B)}{P(B)}$
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