<p>A box contains 12 red and 6 white balls. Balls are drawn from the bag one at a time without replacement. If in 6 draws, there are at least 4 white balls, find the probability that exactly one white ball is drawn in the next two draws. (Binomial coefficients can be left as such.)</p>
Step-by-Step Solution
Key Concept: Use conditional probability: P(exactly 1 white in next 2 | at least 4 white in first 6). The condition 'at least 4 white in first 6 draws' restricts the possible states of remaining balls, which determines probabilities for subsequent draws.
<p><strong>Step 1: Identify the condition</strong> At least 4 white balls in 6 draws means exactly 4, 5, or 6 white balls drawn (and thus 2, 1, or 0 red balls drawn respectively).</p><p><strong>Step 2: Find remaining balls for each case</strong></p><ul><li>Case A: 4W, 2R drawn → Remaining: 8R, 2W (total 10)</li><li>Case B: 5W, 1R drawn → Remaining: 7R, 1W (total 8)</li><li>Case C: 6W, 0R drawn → Remaining: 6R, 0W (total 6)</li></ul><p><strong>Step 3: Calculate probabilities of each case</strong></p><p>P(Case A) = C(6,4)×C(12,2)/C(18,6) = 15×66/18564</p><p>P(Case B) = C(6,5)×C(12,1)/C(18,6) = 6×12/18564</p><p>P(Case C) = C(6,6)×C(12,0)/C(18,6) = 1×1/18564</p><p><strong>Step 4: Find P(exactly 1W in next 2 draws | condition)</strong></p><p>For Case A (8R, 2W): P(1W in 2) = C(2,1)×C(8,1)/C(10,2) = 2×8/45 = 16/45</p><p>For Case B (7R, 1W): P(1W in 2) = C(1,1)×C(7,1)/C(8,2) = 1×7/28 = 1/4</p><p>For Case C (6R, 0W): P(1W in 2) = 0</p><p><strong>Step 5: Apply law of total probability</strong></p><p>P(1W in next 2 | at least 4W in first 6) = [P(A)×(16/45) + P(B)×(1/4) + P(C)×0] / [P(A) + P(B) + P(C)]</p><p>= [990×(16/45) + 72×(1/4)] / [990 + 72 + 1]</p><p>= [352 + 18] / 1063</p><p>= 370/1063</p><p>∴ <strong>Answer: 370/1063</strong></p>
Correct Answer: 370