Question

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 $A$ grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an $A$ grade. Then, the probability that the student is a hostlier is

• A. $\frac{9}{13}$
• B. $\frac{1}{13}$
• C. $\frac{3}{13}$
• D. $\frac{2}{13}$

Answer 1

Answer: (A)

Let $E_1$ : the event that the student is residing in hostel and $E_2$ : the event that the student is not residing in the hostel.
Let $E$ : a student attains $A$ grade,
Then, $E_1$ and $E_2$ are mutually exclusive and exhaustive. Moreover,
$P\left(E_1\right)=60 \%=\frac{60}{100}=\frac{3}{5}$ and $P\left(E_2\right)=40 \%=\frac{40}{100}=\frac{2}{5}$
Then, $ P\left(\frac{E}{E_1}\right)=30 \%=\frac{30}{100}=\frac{3}{10}$
and $P\left(\frac{E}{E_2}\right)=20 \%=\frac{20}{100}=\frac{2}{10}$
By using Bayes' theorem, we get
$P\left(\frac{E_1}{E}\right)=\frac{P\left(\frac{E}{E_1}\right) P\left(E_1\right)}{P\left(\frac{E}{E_1}\right) P\left(E_1\right)+P\left(\frac{E}{E_2}\right) P\left(E_2\right)}$
$=\frac{\frac{3}{10} \times \frac{3}{5}}{\frac{3}{10} \times \frac{3}{5}+\frac{2}{10} \times \frac{2}{5}}$
$=\frac{9}{9+4}=\frac{9}{13}$