<p>Cards are drawn one-by-one at random from a well-shuffled pack of 52 playing cards until 2 aces are obtained from the first time. The probability that 18 draws are required for this is</p>
Step-by-Step Solution
Key Concept: For exactly 18 draws to get 2 aces, the 18th card must be an ace AND exactly 1 ace must appear in the first 17 cards. This transforms the problem into: P(1 ace in first 17 cards) × P(18th card is ace | 1 ace already drawn).
<p><strong>Step 1:</strong> For exactly 18 draws needed, the 18th card must be the 2nd ace. This means exactly 1 ace in first 17 cards and 3 aces remain among last 35 cards (52-17).</p><p><strong>Step 2:</strong> Number of ways to choose 1 ace position in first 17 draws from 4 aces: C(4,1) = 4</p><p>Number of ways to choose 16 non-aces in first 17 draws from 48 non-aces: C(48,16)</p><p>Probability = [C(4,1) × C(48,16) × 1] / [C(52,17) × (52-17)]</p><p><strong>Step 3:</strong> Alternatively, P = [C(4,1) × C(48,16) / C(52,17)] × [3/35]</p><p>= [4 × C(48,16) / C(52,17)] × [3/35]</p><p><strong>Step 4:</strong> Using the formula: P = C(4,1)×C(48,16)/(C(52,18)) × (4)/(52-17) = C(4,1)×C(48,16)×3/(C(52,17)×35)</p><p>Simplifying: P = (4 × 48!×36!)/(17!×31! × 52!) × (3/35) = <strong>12×C(48,16)/(C(52,17)×35)</strong></p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C