Probability
Total Probability
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 9:</b> A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct option(s) with the possible values of \(n_1\) and \(n_2\) is (are)</p>
<p>\(n_1 = 4\) and \(n_2 = 6\)</p>
<p>\(n_1 = 2\) and \(n_2 = 3\)</p>
<p>\(n_1 = 10\) and \(n_2 = 20\)</p>
<p>\(n_1 = 3\) and \(n_2 = 6\)</p>
Step-by-Step Solution
Key Concept: After transferring a ball from box I to box II, the probability of drawing red from box I equals (n₁ - k)/(n₁ + n₂ - 1), where k is 1 if red was transferred, 0 if black was transferred. Use the law of total probability: P(red from I after transfer) = P(red transferred)·P(red|red transferred) + P(black transferred)·P(red|black transferred) = (n₁/(n₁+n₂))·((n₁-1)/(n₁+n₂-1)) + (n₂/(n₁+n₂))·(n₁/(n₁+n₂-1)) = n₁(n₁+n₂-1)/[(n₁+n₂)(n₁+n₂-1)] = n₁/(n₁+n₂) = 1/3.
<p><strong>Step 1:</strong> After transferring one ball from box I to box II, we draw from box I (which now has n₁+n₂-1 balls).</p><p><strong>Step 2:</strong> Use total probability. Let R = event that red ball is drawn from box I after transfer.</p><p>P(R) = P(red transferred)·P(red remains in I | red transferred) + P(black transferred)·P(red in I | black transferred)</p><p>P(R) = (n₁/(n₁+n₂))·((n₁-1)/(n₁+n₂-1)) + (n₂/(n₁+n₂))·(n₁/(n₁+n₂-1))</p><p><strong>Step 3:</strong> Simplify the numerator:</p><p>P(R) = [n₁(n₁-1) + n₂·n₁]/[(n₁+n₂)(n₁+n₂-1)]</p><p>P(R) = [n₁(n₁-1+n₂)]/[(n₁+n₂)(n₁+n₂-1)]</p><p>P(R) = n₁(n₁+n₂-1)/[(n₁+n₂)(n₁+n₂-1)]</p><p>P(R) = n₁/(n₁+n₂) = 1/3</p><p><strong>Step 4:</strong> Therefore: 3n₁ = n₁ + n₂, which gives n₂ = 2n₁</p><p>Any pair (n₁, n₂) satisfying n₂ = 2n₁ where n₁ ≥ 1 is valid.</p><p>∴ Answer: C,D (verify options satisfy n₂ = 2n₁)</p>
Correct Answer: C,D