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A and B are mutually exclusive events of a random experiment and $\mathrm{P}(\mathrm{B}) \neq 1$, then $\mathrm{P}\left(\mathrm{A} \mid \mathrm{B}^{\mathrm{C}}\right)=$
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The correct answer is:
$\frac{\mathrm{P}(\mathrm{A})}{1-\mathrm{P}(\mathrm{B})}$
$\because \mathrm{A}$ and $\mathrm{B}$ are mutually exclusive events.
$\begin{aligned} & \therefore P(A \cap B)=0 \\ & \text { Now, } P\left(A \mid B^c\right)=\frac{P\left(A \cap B^c\right)}{P\left(B^c\right)}=\frac{P(A)-P(A \cap B)}{1-P(B)} \\ & =\frac{P(A)}{1-P(B)}\end{aligned}$
$\begin{aligned} & \therefore P(A \cap B)=0 \\ & \text { Now, } P\left(A \mid B^c\right)=\frac{P\left(A \cap B^c\right)}{P\left(B^c\right)}=\frac{P(A)-P(A \cap B)}{1-P(B)} \\ & =\frac{P(A)}{1-P(B)}\end{aligned}$
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