Probability
Bayes' Theorem
Grade 12
Question:
<p><b>For Problems 8 and 9:</b> Let \(n_1\) and \(n_2\) be the numbers of red and black balls, respectively, in box I. Let \(n_3\) and \(n_4\) be the numbers of red and black balls, respectively, in box II.</p><p><b>Problem 8:</b> One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 1/3, then the correct option(s) with the possible values of \(n_1\), \(n_2\), \(n_3\) and \(n_4\) is (are)</p>
<p>\(n_1 = 3,\ n_2 = 3,\ n_3 = 5,\ n_4 = 15\)</p>
<p>\(n_1 = 3,\ n_2 = 6,\ n_3 = 10,\ n_4 = 50\)</p>
<p>\(n_1 = 8,\ n_2 = 6,\ n_3 = 5,\ n_4 = 20\)</p>
<p>\(n_1 = 6,\ n_2 = 12,\ n_3 = 5,\ n_4 = 20\)</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem: P(Box II | Red) = P(Red | Box II) × P(Box II) / P(Red). The key is that P(Box II | Red) = 1/3 means P(Box I | Red) = 2/3, and since boxes are equally likely initially, the ratio of red ball probabilities from each box must be 2:1.
<p><strong>Step 1:</strong> Apply Bayes' theorem. Given P(Box II | Red) = 1/3, we have P(Box I | Red) = 2/3.</p><p><strong>Step 2:</strong> Using Bayes' theorem: P(Box II | Red) = [P(Red | Box II) × P(Box II)] / P(Red)</p><p>Since P(Box I) = P(Box II) = 1/2, and P(Red) = (1/2)·P(Red | Box I) + (1/2)·P(Red | Box II)</p><p><strong>Step 3:</strong> Let p₁ = n₁/(n₁+n₂) and p₂ = n₃/(n₃+n₄). Then:</p><p>P(Box II | Red) = p₂/(p₁ + p₂) = 1/3</p><p><strong>Step 4:</strong> This gives us p₂ = (1/3)(p₁ + p₂), so 3p₂ = p₁ + p₂, therefore p₁ = 2p₂</p><p><strong>Step 5:</strong> The ratio of red balls in Box I to Box II must satisfy: n₁/(n₁+n₂) = 2 · n₃/(n₃+n₄). Check options against this constraint to identify valid combinations.</p><p>∴ Answer: A,B</p>
Correct Answer: A,B