Question

Three urns contains 2 white and 3 black balls; 3 white and 2 black balls and 4 white and 1 black ball respectively. One ball is drawn from an urn chosen at random and it was found to be white. Find the probability that it was drawn from the first urn.

Answer 1

Let E₁, E₂ and E₃ denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the ball drawn is white.

\[\therefore P\left( E_1 \right) = \frac{1}{3}\]
\[ P\left( E_2 \right) = \frac{1}{3}\]
\[ P\left( E_3 \right) = \frac{1}{3}\]
\[\text{ Now} , \]
\[P\left( A/ E_1 \right) = \frac{2}{5}\]
\[P\left( A/ E_2 \right) = \frac{3}{5}\]
\[P\left( A/ E_3 \right) = \frac{4}{5}\]
\[\text{ Using Bayes' theorem, we get } \]
\[\text{ Required probability }  = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right) + + P\left( E_3 \right)P\left( A/ E_3 \right)}\]
\[ = \frac{\frac{1}{3} \times \frac{2}{5}}{\frac{1}{3} \times \frac{2}{5} + \frac{1}{3} \times \frac{3}{5} + \frac{1}{3} \times \frac{4}{5}}\]
\[ = \frac{2}{2 + 3 + 4} = \frac{2}{9}\]