<p><b>For Problems 6 and 7:</b> A box \(B_1\) contains 1 white ball, 3 red balls, and 2 black balls. Another box \(B_2\) contains 2 white balls, 3 red balls, and 4 black balls. A third box \(B_3\) contains 3 white balls, 4 red balls, and 5 black balls.</p><p><b>Problem 7:</b> If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box \(B_2\) is</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem: P(B₂|1W,1R) = P(1W,1R|B₂)·P(B₂) / P(1W,1R). Calculate the probability of drawing one white and one red ball from each box, then weight by the probability of selecting that box.
<p><strong>Step 1:</strong> Calculate P(1 white, 1 red | each box)</p><p>For B₁ (1W, 3R, 2B, total 6): P(1W,1R) = C(1,1)·C(3,1)·2!/2! / C(6,2) = (1·3)/(15) = 1/5</p><p>For B₂ (2W, 3R, 4B, total 9): P(1W,1R) = C(2,1)·C(3,1)·2! / C(9,2) = (2·3·2)/(36) = 12/36 = 1/3</p><p>For B₃ (3W, 4R, 5B, total 12): P(1W,1R) = C(3,1)·C(4,1)·2! / C(12,2) = (3·4·2)/(66) = 24/66 = 4/11</p><p><strong>Step 2:</strong> Apply Bayes' theorem with P(Bᵢ) = 1/3 for each box</p><p>P(B₂|1W,1R) = P(1W,1R|B₂)·P(B₂) / [P(1W,1R|B₁)·P(B₁) + P(1W,1R|B₂)·P(B₂) + P(1W,1R|B₃)·P(B₃)]</p><p><strong>Step 3:</strong> Calculate denominator = (1/3)·(1/5 + 1/3 + 4/11)</p><p>Common denominator 165: 1/5 = 33/165, 1/3 = 55/165, 4/11 = 60/165</p><p>Sum = 148/165</p><p><strong>Step 4:</strong> P(B₂|1W,1R) = (1/3)·(55/165) / [(1/3)·(148/165)] = 55/148</p><p>∴ Answer: C</p>
Correct Answer: C