A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.

$\frac{5}{9}$

The correct answer is Option (2) → $\frac{5}{9}$ ##

Here, $W_1 = \{4 \text{ white balls}\}$ and $B_1 = \{5 \text{ black balls}\}$

and $W_2 = \{9 \text{ white balls}\}$ and $B_2 = \{7 \text{ black balls}\}$

Let $E_1$ is the event that ball transferred from the first bag is white and $E_2$ is the event that the ball transferred from the first bag is black.

And let $E$ is the event that the ball drawn from the second bag is white.

$∴P(E | E_1) = \frac{10}{17}, P(E | E_2) = \frac{9}{17}$

$\text{and } P(E_1) = \frac{4}{9} \text{ and } P(E_2) = \frac{5}{9}$

$∴P(E) = P(E_1) \cdot P(E | E_1) + P(E_2) \cdot P(E | E_2)$

$= \frac{4}{9} \cdot \frac{10}{17} + \frac{5}{9} \cdot \frac{9}{17} = \frac{40 + 45}{153} = \frac{85}{153} = \frac{5}{9}$