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