Question

Three urns A, B and C contain 6 red and 4 white; 2 red and 6 white; and 1 red and 5 white balls respectively. An urn is chosen at random and a ball is drawn. If the ball drawn is found to be red, find the probability that the ball was drawn from urn A.

Answer 1

Let A, E₁ and E₂ denote the events that the ball is red, bag A is chosen, bag B is chosen and bag C is chosen, respectively.

\[\therefore P\left( E_1 \right) = \frac{1}{3}\]
\[ P\left( E_2 \right) = \frac{1}{3} \]
\[ P\left( E_3 \right) = \frac{1}{3}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{6}{10} = \frac{3}{5}\]
\[P\left( A/ E_2 \right) = \frac{2}{8} = \frac{1}{4}\]
\[P\left( A/ E_3 \right) = \frac{1}{6}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability }  = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + P\left( E_3 \right)P\left( A/ E_3 \right)}\]
\[ = \frac{\frac{1}{3} \times \frac{3}{5}}{\frac{1}{3} \times \frac{3}{5} + \frac{1}{3} \times \frac{1}{4} + \frac{1}{3} \times \frac{1}{6}}\]
\[ = \frac{\frac{3}{5}}{\frac{3}{5} + \frac{1}{4} + \frac{1}{6}} = \frac{36}{61}\]