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Of the students in a college, it is known that $60 \%$ reside in hostel and $40 \%$ are day scholars (not residing in hostel). Previous year results report that $30 \%$ of all students who reside in hostel attain $A$ grade and $20 \%$ of day scholars attain Agrade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an $\mathrm{A}$ grade, what is the probability that the student is a hostlier?
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Verified Answer
Let $\mathrm{E}_1, \mathrm{E}_2$ and $\mathrm{A}$ represents the following:
$\mathrm{E}_1=$ students residing in the hostel,
$\mathrm{E}_2=$ day scholars (not residing in the hostel)
and $A=$ students who attain grade $A$
Now $\quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{60}{100}, \mathrm{P}\left(\mathrm{E}_2\right)=\frac{40}{100}$
$$
\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_1\right)=\frac{30}{100}, \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_2\right)=\frac{20}{100}
$$
Now by Bayes' theorem
$$
\begin{aligned}
&\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right)=\frac{P\left(E_1\right) P\left(A \mid E_1\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)} \\
&=\frac{\frac{60}{100} \times \frac{30}{100}}{\frac{60}{100} \times \frac{30}{100}+\frac{40}{100} \times \frac{20}{100}}=\frac{9}{13}
\end{aligned}
$$
$\mathrm{E}_1=$ students residing in the hostel,
$\mathrm{E}_2=$ day scholars (not residing in the hostel)
and $A=$ students who attain grade $A$
Now $\quad \mathrm{P}\left(\mathrm{E}_1\right)=\frac{60}{100}, \mathrm{P}\left(\mathrm{E}_2\right)=\frac{40}{100}$
$$
\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_1\right)=\frac{30}{100}, \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_2\right)=\frac{20}{100}
$$
Now by Bayes' theorem
$$
\begin{aligned}
&\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right)=\frac{P\left(E_1\right) P\left(A \mid E_1\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)} \\
&=\frac{\frac{60}{100} \times \frac{30}{100}}{\frac{60}{100} \times \frac{30}{100}+\frac{40}{100} \times \frac{20}{100}}=\frac{9}{13}
\end{aligned}
$$
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