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:

Box I contains 3 black and 4 white balls (7 total).

Box II contains 2 black and 2 white balls (4 total).

Box III contains 4 black and 3 white balls (7 total).

A box is selected at random, so each box has probability $\frac{1}{3}$ of being selected.

The probability of drawing a black ball from each box:

From Box I: $P(\text{Black | Box I}) = \frac{3}{7}$

From Box II: $P(\text{Black | Box II}) = \frac{2}{4} = \frac{1}{2}$

From Box III: $P(\text{Black | Box III}) = \frac{4}{7}$

The total probability of drawing a black ball:

$P(\text{Black}) = P(\text{Box I}) \times P(\text{Black | Box I}) + P(\text{Box II}) \times P(\text{Black | Box II}) + P(\text{Box III}) \times P(\text{Black | Box III})$

$P(\text{Black}) = \frac{1}{3} \times \frac{3}{7} + \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{4}{7}$

$P(\text{Black}) = \frac{3}{21} + \frac{1}{6} + \frac{4}{21}$

Converting to common denominator 42:

$P(\text{Black}) = \frac{6}{42} + \frac{7}{42} + \frac{8}{42}$

$P(\text{Black}) = \frac{21}{42}$

$P(\text{Black}) = \frac{1}{2}$

Using Bayes' theorem to find the probability that the black ball came from Box III:

$P(\text{Box III | Black}) = \frac{P(\text{Box III and Black})}{P(\text{Black})}$

$P(\text{Box III and Black}) = P(\text{Box III}) \times P(\text{Black | Box III})$

$P(\text{Box III and Black}) = \frac{1}{3} \times \frac{4}{7}$

$P(\text{Box III and Black}) = \frac{4}{21}$

Therefore:

$P(\text{Box III | Black}) = \frac{\frac{4}{21}}{\frac{1}{2}}$

$P(\text{Box III | Black}) = \frac{4}{21} \times \frac{2}{1}$

$P(\text{Box III | Black}) = \frac{8}{21}$