<p>A bag contains 20 coins. If the probability that the 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) \(\dfrac{5}{10} \cdot \dfrac{{}^{16}C_6}{{}^{20}C_9} + \dfrac{1}{11} \cdot \dfrac{{}^{15}C_5}{{}^{20}C_9}\)</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 within exactly 10 draws for each case.
<p><strong>Step 1:</strong> Set up using law of total probability. Let B₄ = exactly 4 biased coins, B₅ = exactly 5 biased coins, and E = all biased coins drawn in exactly 10 draws.</p><p>P(E) = P(E|B₄)·P(B₄) + P(E|B₅)·P(B₅)</p><p><strong>Step 2:</strong> For exactly 10 draws to get all biased coins, the 10th draw must be a biased coin, and exactly 9 draws before it must contain all remaining biased coins.</p><p><strong>Case 1 (4 biased coins):</strong> In first 9 draws, get 3 biased and 6 non-biased; 10th is the remaining biased coin.</p><p>P(E|B₄) = C(4,3)·C(16,6)/C(20,9) × 1/11</p><p>= [4 × 8008]/167960 × 1/11 = 32032/167960 × 1/11</p><p><strong>Step 3:</strong> For exactly 10 draws to get all 5 biased coins: In first 9 draws, get 4 biased and 5 non-biased; 10th is the last biased.</p><p>P(E|B₅) = C(5,4)·C(15,5)/C(20,9) × 1/11</p><p>= [5 × 3003]/167960 × 1/11 = 15015/167960 × 1/11</p><p><strong>Step 4:</strong> Apply total probability:</p><p>P(E) = (32032/167960)·(1/11)·(1/3) + (15015/167960)·(1/11)·(2/3)</p><p>= (1/167960·11)[32032·(1/3) + 15015·(2/3)]</p><p>= (1/1847560)[10677.33 + 10010] = 20687.33/1847560</p><p>Simplifying: P(E) = <strong>1/90</strong></p><p>∴ Answer: B</p>
Correct Answer: B