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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{8}{21}\)
Let \(E_1, E_2, E_3\) be the events that box I, II or III is selected, and A the event that the ball drawn is black. Each box is equally likely:
\( \displaystyle P(E_1) = P(E_2) = P(E_3) = \frac{1}{3} \)
\( \displaystyle P(A|E_1) = \frac{3}{7},\quad P(A|E_2) = \frac{2}{4} = \frac{1}{2},\quad P(A|E_3) = \frac{4}{7} \)
By Bayes’ theorem,
\( \displaystyle P(E_3|A) = \frac{P(E_3)P(A|E_3)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)} \)
The common factor \(\frac{1}{3}\) cancels:
\( \displaystyle = \frac{\frac{4}{7}}{\frac{3}{7} + \frac{1}{2} + \frac{4}{7}} = \frac{\frac{4}{7}}{\frac{3}{2}} \)
\( \displaystyle = \frac{4}{7} \times \frac{2}{3} = \frac{8}{21} \)