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Question
A bag A contains 2 white and 3 red balls and a bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag B.
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Solution
Let A, E1 and E2 denote the events that the ball is red, bag A is chosen and bag B is chosen, respectively.
\[\therefore P\left( E_1 \right) = \frac{1}{2} \]
\[ P\left( E_2 \right) = \frac{1}{2}\]
\[\text{Now} , \]
\[P\left( A/ E_1 \right) = \frac{3}{5}\]
\[P\left( A/ E_2 \right) = \frac{5}{9}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_2 /A \right) = \frac{P\left( E_2 \right)P\left( A/ E_2 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{5}{9}}\]
\[ = \frac{25}{52}\]
\[ P\left( E_2 \right) = \frac{1}{2}\]
\[\text{Now} , \]
\[P\left( A/ E_1 \right) = \frac{3}{5}\]
\[P\left( A/ E_2 \right) = \frac{5}{9}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_2 /A \right) = \frac{P\left( E_2 \right)P\left( A/ E_2 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{5}{9}}\]
\[ = \frac{25}{52}\]
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