Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel).
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, what is the probability that the student is hostler?

Solution:

Let E1 and E2 be the events that the student is a hostler and a day scholar respectively and A be the event that the chosen student gets grade A.

$\therefore \mathrm{P}\left(\mathrm{E}_{1}\right)=60 \%=\frac{60}{100}=0.6$

$\mathrm{P}\left(\mathrm{E}_{2}\right)=40 \%=\frac{40}{100}=0.4$

$P\left(A \mid E_{1}\right)=P($ student getting an $A$ grade is a hostler $)=30 \%=0.3$

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\mathrm{P}$ (student getting an A grade is a day scholar $)=20 \%=0.2$

The probability that a randomly chosen student is a hostler, given that he has an $\mathrm{A}$ grade, is given by $\mathrm{P}\left(\mathrm{E}_{\mid} \mid \mathrm{A}\right)$.

By using Bayes’ theorem, we obtain’

$P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)}$

$=\frac{0.6 \times 0.3}{0.6 \times 0.3+0.4 \times 0.2}$

$=\frac{0.18}{0.26}$

$=\frac{18}{26}$

$=\frac{9}{13}$

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