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.

$\frac{1}{3}$

The correct answer is Option (3) → $\frac{1}{3}$ ##

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

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

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

$∴P(U_1) = P(U_2) = P(U_3) = \frac{1}{3}$

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(E_1) = P(E_2) = P(E_3) = 1/3$

Now, let $E$ be the event that white ball is drawn.

$∴P(E | E_1) = \frac{2}{5}, P(E | E_2) = \frac{3}{5}, P(E | E_3) = \frac{4}{5}$

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

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

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