<p>A box '\(A\)' contains 2 white, 3 red and 2 black balls. Another box '\(B\)' contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box '\(B\)' is</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem: find P(Box B | one white, one red) by calculating the probability of drawing one white and one red from each box, then apply the conditional probability formula with equal prior probabilities for box selection.
<p><strong>Step 1:</strong> Calculate P(1 white, 1 red | Box A)</p><p>Box A: 2 white, 3 red, 2 black (total 7 balls)</p><p>P(1W, 1R | A) = (2/7 × 3/6) + (3/7 × 2/6) = 6/42 + 6/42 = 12/42 = 2/7</p><p><strong>Step 2:</strong> Calculate P(1 white, 1 red | Box B)</p><p>Box B: 4 white, 2 red, 3 black (total 9 balls)</p><p>P(1W, 1R | B) = (4/9 × 2/8) + (2/9 × 4/8) = 8/72 + 8/72 = 16/72 = 2/9</p><p><strong>Step 3:</strong> Apply Bayes' theorem</p><p>P(Box B | 1W, 1R) = [P(1W, 1R | B) × P(B)] / [P(1W, 1R | A) × P(A) + P(1W, 1R | B) × P(B)]</p><p>Since P(A) = P(B) = 1/2:</p><p>P(Box B | 1W, 1R) = (2/9 × 1/2) / [(2/7 × 1/2) + (2/9 × 1/2)]</p><p>= (2/9) / (2/7 + 2/9) = (2/9) / [(18 + 14)/63] = (2/9) / (32/63)</p><p>= (2/9) × (63/32) = 126/288 = 7/16</p><p>∴ Answer: B</p>
Correct Answer: B