40% students of a college reside in hostel and the remaining reside outside.At the end of year,50% of the hosteliers got A grade while from outside students,only 30% got A grade in the examination.At the end of the year,a student of the college was chosen at random and was found to got A grade.What is the probability that the selected student was a hostelier?

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Solution

Let $E_{1}$ :Student resides in the hostel, $E_{2}$ :Student resides outside the hostel,A:Getting A grade in the examination.
$P (E_{1}) = \frac{40}{100} = \frac{2}{5}, P (E_{2}) = \frac{3}{5}, P (A | E_{1}) = \frac{50}{100} = \frac{1}{2}, P (A | E_{2}) = \frac{30}{100} = \frac{3}{10}$
By Baye's theorem, $P (E_{1} | A) = \frac{P (A | E_{1}) P (E_{1})}{P (A | E_{1}) P (E_{1}) + P (A | E_{2}) P (E_{2})} = \frac{\frac{2}{5} \times \frac{1}{2}}{\frac{2}{5} \times \frac{1}{2} + \frac{3}{5} \times \frac{3}{10}}$
$∴ P (E_{1} | A) = \frac{10}{19}$ .

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Let E1:Student resides in the hostel,E2:Student resides outside the hostel,A:Getting A grade in the examination. P(E1)=40100=25,P(E2)=35,P(A|E1)=50100=12,P(A|E2)=30100=310 By Baye's theorem,P(E1|A)=P(A|E1)P(E1)P(A|E1)P(E1)+P(A|E2)P(E2)=25×1225×12+35×310 ∴ P(E1|A)=1019.