Question

Three boxes \( B_1, B_2, B_3 \) contain white, black, and red balls as follows: \[ \begin{array}{|c|c|c|c|} \hline & \textbf{White} & \textbf{Black} & \textbf{Red} \\ \hline B_1 & 2 & 1 & 2 \\ B_2 & 3 & 2 & 4 \\ B_3 & 4 & 3 & 2 \\ \hline \end{array} \] A die is thrown:
- Box \( B_1 \) is chosen if the die shows 1 or 2
- Box \( B_2 \) is chosen if the die shows 3 or 4
- Box \( B_3 \) is chosen if the die shows 5 or 6
A ball is drawn at random from the selected box. the ball drawn is red, find the probability it came from box \( B_2 \):

• A. \( \frac{7}{12} \)
• B. \( \frac{5}{12} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{3}{26} \)

Hint:

Use Bayes’ Theorem with case-wise total probability to solve reverse conditional problems.

Answer 1

The Correct Option is B

\[ P(B_1) = P(B_2) = P(B_3) = \frac{1}{3} \] Probability of red from: - \( B_1 \): \( \frac{2}{5} \) - \( B_2 \): \( \frac{4}{9} \) - \( B_3 \): \( \frac{2}{9} \) Total probability of red: \[ P(R) = \frac{1}{3} \cdot \frac{2}{5} + \frac{1}{3} \cdot \frac{4}{9} + \frac{1}{3} \cdot \frac{2}{9} = \frac{2}{15} + \frac{4}{27} + \frac{2}{27} = \frac{2}{15} + \frac{6}{27} = \frac{2}{15} + \frac{2}{9} = \frac{4 + 10}{45} = \frac{14}{45} \] Now: \[ P(B_2 | R) = \frac{P(B_2) \cdot P(R | B_2)}{P(R)} = \frac{\frac{1}{3} \cdot \frac{4}{9}}{\frac{14}{45}} = \frac{4}{27} \cdot \frac{45}{14} = \frac{180}{378} = \frac{5}{12} \]