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Question
Of the students in a college, it is known that 60% reside in a hostel and 40% do not reside in hostel. Previous year results report that 30% of students residing in hostel attain A grade and 20% of ones not residing in hostel attain A grade in their annual examination. At the end of the year, one students is chosen at random from the college and he has an A grade. What is the probability that the selected student is a hosteler?
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Solution
Let A, E1 and E2 denote the events that the selected student attains grade A, resides in a hostel and does not reside in a hostel, respectively.
\[\therefore P\left( E_1 \right) = \frac{60}{100} \]
\[ P\left( E_2 \right) = \frac{40}{100}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{30}{100}\]
\[P\left( A/ E_2 \right) = \frac{20}{100}\]
\[\text{ Using Bayes' theorem, we get} \]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{60}{100} \times \frac{30}{100}}{\frac{60}{100} \times \frac{30}{100} + \frac{40}{100} \times \frac{20}{100}}\]
\[ = \frac{18}{18 + 8} = \frac{9}{13}\]
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