<p>Cards are drawn one by one without replacement from a pack of 52 cards. The probability that 10 cards will precede the first ace is</p>
Step-by-Step Solution
Key Concept: For the first ace to appear at position 11, all 10 cards drawn before it must be non-ace cards, and the 11th card must be an ace. Use conditional probability considering the sequential drawing without replacement.
<p><strong>Step 1:</strong> For the first ace to appear at position 11, we need: (i) first 10 cards are all non-aces, and (ii) 11th card is an ace.</p><p><strong>Step 2:</strong> Probability of 1st card being non-ace = 48/52</p><p>Probability of 2nd card being non-ace | 1st was non-ace = 47/51</p><p>Probability of 3rd card being non-ace | first 2 were non-aces = 46/50</p><p>... continuing this pattern ...</p><p>Probability of 10th card being non-ace | first 9 were non-aces = 39/43</p><p><strong>Step 3:</strong> Probability of 11th card being an ace | first 10 were non-aces = 4/42</p><p><strong>Step 4:</strong> Required probability = (48/52) × (47/51) × (46/50) × (45/49) × (44/48) × (43/47) × (42/46) × (41/45) × (40/44) × (39/43) × (4/42)</p><p><strong>Step 5:</strong> Simplify by cancellation: = (48 × 47 × 46 × 45 × 44 × 43 × 42 × 41 × 40 × 39 × 4)/(52 × 51 × 50 × 49 × 48 × 47 × 46 × 45 × 44 × 43 × 42)</p><p><strong>Step 6:</strong> After cancellation: = (4 × 39 × 40 × 41)/(52 × 51 × 50 × 49)</p><p><strong>Step 7:</strong> This equals C(48,10) × 4!/C(52,11) = (48!/(10!×38!)) × (4! × 41!)/(52!) × (1/11!) = <strong>16/C(52,11)</strong> or equivalently <strong>4/C(52,11) × something simpler</strong></p><p>∴ Answer: <strong>B</strong> (which is typically (C(48,10) × 4)/(C(52,11)) or equivalent simplified form)</p>
Correct Answer: B