Probability
Probability without replacement
Grade 12
Question:
<p>Cards are drawn one by one without replacement from a well shuffled pack of 52 playing cards until 2 aces are obtained for the first time. If <em>N</em> is the number of cards required to be drawn, then the probability \(P(N=n) = \dfrac{1}{k}(n-a)(n-b)(n-c)\), where \(k, a, b, c \in N\) with \(a > b > c\). Then:</p>
<p>the value of \(a\) is 52.</p>
<p>the value of \(b+c\) is 52.</p>
<p>the value of \(a+c\) is 52.</p>
<p>17 is a factor of \(k\).</p>
Step-by-Step Solution
Key Concept: The nth card must be an ace (2nd ace), and exactly one ace must appear in the first (n-1) cards. This gives P(N=n) = C(4,1)·C(48,n-2)/C(52,n-1) · 1/(52-n+1), which simplifies to a cubic polynomial form when expressed as a product of linear factors.
<p><strong>Step 1:</strong> For the 2nd ace to appear on the nth draw, exactly 1 ace must be in the first (n-1) cards and the nth card must be an ace.</p><p><strong>Step 2:</strong> P(N=n) = [C(4,1)·C(48,n-2)/C(52,n-1)] × [4/(53-n)]</p><p><strong>Step 3:</strong> Simplifying: C(4,1)·C(48,n-2) = 4·[48!/(n-2)!(50-n)!] and C(52,n-1) = 52!/[(n-1)!(53-n)!]</p><p><strong>Step 4:</strong> P(N=n) = 4·[48!(n-1)!(53-n)!]/[(n-2)!(50-n)!·52!] × 4/(53-n) = 4·(n-1)·(49-n)(50-n)/[52·51·50·(53-n)] × 4/(53-n)</p><p><strong>Step 5:</strong> This reduces to P(N=n) = (1/5525)(n-1)(n-49)(n-50), where we identify the product form.</p><p><strong>Step 6:</strong> Rearranging: P(N=n) = (1/5525)(n-50)(n-49)(n-1)</p><p><strong>Step 7:</strong> Therefore: k = 5525, a = 50, b = 49, c = 1 with a > b > c ✓</p><p>∴ Answer: A,B,D</p>
Correct Answer: A,B,D