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Two die are rolled. If $\mathrm{A}$ denote the event that the same number shows on each die and B denote the event that the sum of the numbers on both dice is greater than 7, then $\mathrm{P}(\mathrm{A} \mid \mathrm{B})$ and $\mathrm{P}(\mathrm{B} \mid \mathrm{A})$ respectively are
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The correct answer is:
$\frac{1}{5}, \frac{1}{2}$
Since $A=\{(1,1),(2,2),(3,3),(4,4),(5,5)$,$(6,6)\}$
$\begin{aligned}
& \text { and } B=\left\{\begin{array}{l}
(2,6),(3,5),(3,6),(4,4),(4,5), \\
(4,6),(5,3),(5,4),(5,5),(5,6) \\
(6,2),(6,3),(6,4),(6,5),(6,6)
\end{array}\right\} \\
& A \cap B=\{(4,4),(5,5),(6,6)\} \\
& \text { Now, } P(A)=\frac{6}{36}=\frac{1}{6}, P(B)=\frac{15}{36}=\frac{5}{12} \\
& P(A \cap B)=\frac{3}{36}=\frac{1}{12}
\end{aligned}$
$\begin{aligned} & P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{12}}{\frac{5}{12}}=\frac{1}{5} \\ & P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}\end{aligned}$
$\begin{aligned}
& \text { and } B=\left\{\begin{array}{l}
(2,6),(3,5),(3,6),(4,4),(4,5), \\
(4,6),(5,3),(5,4),(5,5),(5,6) \\
(6,2),(6,3),(6,4),(6,5),(6,6)
\end{array}\right\} \\
& A \cap B=\{(4,4),(5,5),(6,6)\} \\
& \text { Now, } P(A)=\frac{6}{36}=\frac{1}{6}, P(B)=\frac{15}{36}=\frac{5}{12} \\
& P(A \cap B)=\frac{3}{36}=\frac{1}{12}
\end{aligned}$
$\begin{aligned} & P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{12}}{\frac{5}{12}}=\frac{1}{5} \\ & P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{\frac{1}{12}}{\frac{1}{6}}=\frac{1}{2}\end{aligned}$
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