Question

Urn A contains 6 white and 2 black balls; urn B contains 5 white and 3 black balls and urn C contains 4 white and 4 black balls. If an urn is chosen at random and a ball is drawn at random from it, then the probability that the ball drawn is white is

• A. \( \frac{3}{8} \)
• B. \( \frac{5}{8} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{3}{4} \)

Hint:

If the total number of balls in each urn is the same, the total probability is simply the average of the number of white balls divided by the total per urn.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

This problem uses the Law of Total Probability. We choose an urn first, then choose a ball from that urn.
Step 2: Key Formula or Approach:

\[ P(W) = P(A)P(W|A) + P(B)P(W|B) + P(C)P(W|C) \] where \( P(A), P(B), P(C) \) are probabilities of choosing the urns, and \( P(W|X) \) is the probability of drawing a white ball from urn X.
Step 3: Detailed Explanation:

1. Probability of choosing any urn: \( P(A) = P(B) = P(C) = \frac{1}{3} \). 2. Probabilities of White balls: - Urn A: 6 White, 2 Black. Total = 8. \( P(W|A) = \frac{6}{8} \). - Urn B: 5 White, 3 Black. Total = 8. \( P(W|B) = \frac{5}{8} \). - Urn C: 4 White, 4 Black. Total = 8. \( P(W|C) = \frac{4}{8} \). 3. Total Probability: \[ P(W) = \frac{1}{3} \times \frac{6}{8} + \frac{1}{3} \times \frac{5}{8} + \frac{1}{3} \times \frac{4}{8} \] \[ = \frac{1}{3} \left( \frac{6}{8} + \frac{5}{8} + \frac{4}{8} \right) \] \[ = \frac{1}{3} \left( \frac{15}{8} \right) = \frac{5}{8} \]
Step 4: Final Answer:

The probability is \( \frac{5}{8} \).