Let box I contains 3 black and 4 white balls, box II contains 2 black and 2 white balls, box III contains 4 black and 3 white balls. A box is selected at random and then a ball is randomly drawn from the selected box. If the color of the ball is black then the probability that the ball is drawn from box III, is:

$\frac{8}{21}$

The correct answer is Option (3) → $\frac{8}{21}$

Total boxes = 3. Probability of selecting any box = $\frac{1}{3}$.

Define events:

$B_1$: Ball drawn from Box I

$B_2$: Ball drawn from Box II

$B_3$: Ball drawn from Box III

$A$: Black ball is drawn

Now, compute:

$P(A|B_1) = \frac{3}{7}$, $P(A|B_2) = \frac{2}{4} = \frac{1}{2}$, $P(A|B_3) = \frac{4}{7}$

By Bayes' Theorem:

$P(B_3|A) = \frac{P(B_3) \cdot P(A|B_3)}{P(B_1)P(A|B_1) + P(B_2)P(A|B_2) + P(B_3)P(A|B_3)}$

$= \frac{\frac{1}{3} \cdot \frac{4}{7}}{\frac{1}{3} \cdot \frac{3}{7} + \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{4}{7}}$

Numerator: $\frac{4}{21}$

Denominator: $\frac{3}{21} + \frac{1}{6} + \frac{4}{21} = \frac{7}{21} + \frac{1}{6} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$

$P(B_3|A) = \frac{4/21}{1/2} = \frac{8}{21}$