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If $A$ and $B$ are mutually exclusive events with $P(B) \neq 1$, then $P(A \mid \bar{B})$ is equal to
(Here $\bar{B}$ is the complement of the event $B$ )
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(Here $\bar{B}$ is the complement of the event $B$ )
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The correct answer is:
$\frac{P(A)}{1-P(B)}$
Given, $A$ and $B$ are mutually exclusive events So, $(A \cap B)=\phi$.
Now, $\begin{aligned} P(A \mid \bar{B}) & =\frac{P(A \cap \bar{B})}{P(\bar{B})}=\frac{P(A)-P(A \cap B)}{P(\bar{B})} \\ & =\frac{P(A)}{1-P(B)}\end{aligned}$
Now, $\begin{aligned} P(A \mid \bar{B}) & =\frac{P(A \cap \bar{B})}{P(\bar{B})}=\frac{P(A)-P(A \cap B)}{P(\bar{B})} \\ & =\frac{P(A)}{1-P(B)}\end{aligned}$
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