There are two boxes, namely box-I and box-II. Box-I contains 3 red and 6 black balls. Box-II contains 5 red and 5 black balls. One of the two boxes is selected at random and a ball is drawn at random. The ball drawn is found to be red. Find the probability that this red ball comes from box-II.

$\frac{3}{5}$

The correct answer is Option (2) → $\frac{3}{5}$ ##

Let $E_1 = \text{Selecting Box-I}$

$E_2 = \text{Selecting Box-II}$

$A = \text{getting a red ball from the selected box}$

Here, $P(E_1) = \frac{1}{2}, P(E_2) = \frac{1}{2}$

$P(A/E_1) = \frac{3}{9} = \frac{1}{3}$

$P(A/E_2) = \frac{5}{10} = \frac{1}{2}$

Required probability = $P\left(\frac{E_2}{A}\right)$

= [Probability that Red ball comes out from Box-II]

Using Bayes' theorem,

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

$= \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{6} + \frac{1}{4}}$

$= \frac{\frac{1}{4}}{\frac{4 + 6}{24}} = \frac{1}{4} \cdot \frac{24}{10} = \frac{6}{10} = \frac{3}{5}$

Thus, the probability that the red ball comes out from Box-II is $\frac{3}{5}$.