<p>A bag contains 20 coins. If the probability that bag contains exactly 4 biased coin is 1/3 and that of exactly 5 biased coin is 2/3, then the probability that all the biased coin are sorted out from the bag in exactly 10 draws is</p>
<p>(1) \(\dfrac{5}{33} \cdot \dfrac{{}^{16}C_6}{{}^{20}C_9} + \dfrac{1}{11} \cdot \dfrac{{}^{15}C_5}{{}^{20}C_9}\)</p>
<p>(2) \(\dfrac{2}{33}\left[\dfrac{{}^{16}C_6 + 5 \cdot {}^{15}C_5}{{}^{20}C_9}\right]\)</p>
<p>(3) \(\dfrac{5}{33} \cdot \dfrac{{}^{16}C_7}{{}^{20}C_9} + \dfrac{1}{11} \cdot \dfrac{{}^{15}C_6}{{}^{20}C_9}\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Use the law of total probability by conditioning on the number of biased coins (4 or 5), then calculate the probability of drawing all biased coins in exactly 10 draws from a finite population.
<p><strong>Step 1:</strong> Set up using law of total probability. Let B₄ = event of exactly 4 biased coins, B₅ = event of exactly 5 biased coins.</p><p>P(all biased sorted in 10 draws) = P(A|B₄)·P(B₄) + P(A|B₅)·P(B₅)</p><p><strong>Step 2:</strong> For exactly 4 biased coins: To draw all 4 biased coins in exactly 10 draws, we need all 4 biased coins plus 6 unbiased coins from 16 unbiased coins.</p><p>P(A|B₄) = C(4,4)·C(16,6)/C(20,10) = C(16,6)/C(20,10)</p><p><strong>Step 3:</strong> For exactly 5 biased coins: To draw all 5 biased coins in exactly 10 draws, we need all 5 biased coins plus 5 unbiased coins from 15 unbiased coins.</p><p>P(A|B₅) = C(5,5)·C(15,5)/C(20,10) = C(15,5)/C(20,10)</p><p><strong>Step 4:</strong> Calculate:</p><p>C(16,6) = 8008, C(15,5) = 3003, C(20,10) = 184756</p><p>P(A|B₄) = 8008/184756, P(A|B₅) = 3003/184756</p><p><strong>Step 5:</strong> Apply total probability:</p><p>P(A) = (8008/184756)·(1/3) + (3003/184756)·(2/3)</p><p>P(A) = (8008 + 6006)/(3·184756) = 14014/554268 = 1/39.5 ≈ <strong>1/39 or equivalent simplified form</strong></p><p>∴ Answer: B</p>
Correct Answer: B