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If $\mathrm{P}(\mathrm{A})=\frac{1}{12}, \mathrm{P}(\mathrm{B})=\frac{5}{12}$ and $\mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{1}{15}$ then $\mathrm{p}(\mathrm{A} \cup \mathrm{B})$ is equal to
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Verified Answer
The correct answer is:
$\frac{89}{180}$
Given
$\mathrm{P}(\mathrm{A})=\frac{1}{12}, \mathrm{P}(\mathrm{B})=\frac{5}{12}, \mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{1}{15}$
We know that $P(B / A)=\frac{P(A \cap B)}{P(A)}$
$\begin{array}{l}
\Rightarrow \frac{1}{15}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{1 / 12} \\
\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{15 \times 12}=\frac{1}{180}
\end{array}$
But,
$\begin{array}{l}
\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
\Rightarrow \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{1}{12}+\frac{5}{12}-\frac{1}{180}=\frac{89}{180}
\end{array}$
$\mathrm{P}(\mathrm{A})=\frac{1}{12}, \mathrm{P}(\mathrm{B})=\frac{5}{12}, \mathrm{P}(\mathrm{B} / \mathrm{A})=\frac{1}{15}$
We know that $P(B / A)=\frac{P(A \cap B)}{P(A)}$
$\begin{array}{l}
\Rightarrow \frac{1}{15}=\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{1 / 12} \\
\Rightarrow \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{15 \times 12}=\frac{1}{180}
\end{array}$
But,
$\begin{array}{l}
\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
\Rightarrow \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{1}{12}+\frac{5}{12}-\frac{1}{180}=\frac{89}{180}
\end{array}$
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