Bag $A$ contains 3 red and 2 black balls, while bag $B$ contains 2 red and 3 black balls. A ball drawn at random from bag $A$ is transferred to bag $B$ and then one ball is drawn at random from bag $B$. If this ball was found to be a red ball, find the probability that the ball drawn from bag $A$ was red.

$\frac{9}{13}$

The correct answer is Option (2) → $\frac{9}{13}$ ##

Let the events be:

$E_1$: Transferring a red ball from $A$ to $B$

$E_2$: Transferring a black ball from $A$ to $B$

$A$: Getting a red ball from bag $B$

$P(E_1) = \frac{3}{5}, P(E_2) = \frac{2}{5}$

$P(A/E_1) = \frac{3}{6} = \frac{1}{2}, P(A/E_2) = \frac{2}{6} = \frac{1}{3}$

$P(E_1/A) = \frac{P(E_1) \cdot P(A/E_1)}{P(E_1) \cdot P(A/E_1) + P(E_2) \cdot P(A/E_2)}$

$= \frac{\frac{3}{5} \cdot \frac{1}{2}}{\frac{3}{5} \cdot \frac{1}{2} + \frac{2}{5} \cdot \frac{1}{3}} = \frac{9}{13}$