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Question
A bag contains 1 white and 6 red balls, and a second bag contains 4 white and 3 red balls. One of the bags is picked up at random and a ball is randomly drawn from it, and is found to be white in colour. Find the probability that the drawn ball was from the first bag.
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Solution
Let A, E1 and E2 denote the events that the ball is white, bag I is chosen and bag II is chosen, respectively.
\[\therefore P\left( E_1 \right) = \frac{1}{2} \]
\[ P\left( E_2 \right) = \frac{1}{2}\]
\[Now, \]
\[P\left( A/ E_1 \right) = \frac{1}{7}\]
\[P\left( A/ E_2 \right) = \frac{4}{7}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability } = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{2} \times \frac{1}{7}}{\frac{1}{2} \times \frac{1}{7} + \frac{1}{2} \times \frac{4}{7}}\]
\[ = \frac{1}{1 + 4} = \frac{1}{5}\]
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