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Let $A$ and $B$ be not mutually exclusive events. If $P(A)=\frac{4}{9}, P(A \cap \bar{B})=\frac{3}{7}$ then $P\left(\frac{B}{A}\right)=$
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The correct answer is:
$\frac{1}{28}$
We have,
$P(A)=\frac{4}{9}, P(A \cap \bar{B})=\frac{3}{7}$
$\begin{aligned} & \text { Now, } P(A \cap \bar{B})=P(A)-P(A \cap B) \\ & \Rightarrow \quad P(A \cap B)=P(A)-P(A \cap \bar{B}) \\ & =\frac{4}{9}-\frac{3}{7}=\frac{28-27}{63}=\frac{1}{63} \\ & \therefore P(B / A)=\frac{P(B \cap A)}{P(A)}=\frac{\left(\frac{1}{63}\right)}{\left(\frac{4}{9}\right)}=\frac{1}{28} \\ & \end{aligned}$
$P(A)=\frac{4}{9}, P(A \cap \bar{B})=\frac{3}{7}$
$\begin{aligned} & \text { Now, } P(A \cap \bar{B})=P(A)-P(A \cap B) \\ & \Rightarrow \quad P(A \cap B)=P(A)-P(A \cap \bar{B}) \\ & =\frac{4}{9}-\frac{3}{7}=\frac{28-27}{63}=\frac{1}{63} \\ & \therefore P(B / A)=\frac{P(B \cap A)}{P(A)}=\frac{\left(\frac{1}{63}\right)}{\left(\frac{4}{9}\right)}=\frac{1}{28} \\ & \end{aligned}$
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