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If $A$ and $B$ be independent events with $P(A)=\frac{1}{3}$ and $P(B)=\frac{2}{7}$, then the value of $P\left(\frac{A}{B^C}\right)$ is
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The correct answer is:
$\frac{1}{3}$
Given, $P(A)=\frac{1}{3}, P(B)=\frac{2}{7}$
$P\left(A / B^{\prime}\right)=\frac{P\left(A \cap B^{\prime}\right)}{P\left(B^{\prime}\right)}=\frac{P(A)-P(A \cap B)}{1-P(B)}$
$=\frac{P(A)-P(A) P(B)}{1-P(B)}$ $[\because A$ and $B$ are independent events $]$
$=\frac{\frac{1}{3}-\frac{1}{3} \times \frac{2}{7}}{1-2 / 7}=\frac{(7-2)}{3(7-2)}=\frac{1}{3}$
$P\left(A / B^{\prime}\right)=\frac{P\left(A \cap B^{\prime}\right)}{P\left(B^{\prime}\right)}=\frac{P(A)-P(A \cap B)}{1-P(B)}$
$=\frac{P(A)-P(A) P(B)}{1-P(B)}$ $[\because A$ and $B$ are independent events $]$
$=\frac{\frac{1}{3}-\frac{1}{3} \times \frac{2}{7}}{1-2 / 7}=\frac{(7-2)}{3(7-2)}=\frac{1}{3}$
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