There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.

Let ${U_1} = \{2 white, 3 black balls \}$

${U_2} = \{ 3$ white, 2 black balls$\}$

and
${U_3} = \{ 4$ white, 1 black balls $\}$

$\therefore$ $P\left( {{U_1}} \right) = P\left( {{U_2}} \right) = P\left( {{U_3}} \right) = \frac{1}{3}$

Let ${E_1}$ be the event that a ball is chosen from urn ${U_1},{E_2}$

be the event that a ball is chosen from urn ${U_2}$ and ${E_3}$

be the event that a ball is chosen from urn ${U_3}$.

Also, $P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = P\left( {{E_3}} \right) = 1/3$

Now, let $E$ be the event that white ball is drawn.
$\therefore$ $P\left( {E/{E_1}} \right) = \frac{2}{5},P\left( {E/{E_2}} \right) = \frac{3}{5},P\left( {E/{E_3}} \right) = \frac{4}{5}$

Now, $P\left( {{E_2}/E} \right) = \frac{{P\left( {{E_2}} \right) \cdot P\left( {E/{E_2}} \right)}}{{P\left( {{E_1}} \right) \cdot P\left( {E/{E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {E/{E_2}} \right) + P\left( {{E_3}} \right) \cdot P\left( {E/{E_3}} \right)}}$

$= \frac{{\frac{1}{3} \cdot \frac{3}{5}}}{{\frac{1}{3} \cdot \frac{2}{5} + \frac{1}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{4}{5}}}$

$= \frac{{\frac{3}{{15}}}}{{\frac{2}{{15}} + \frac{3}{{15}} + \frac{4}{{15}}}} = \frac{3}{9} = \frac{1}{3}$