Probability
Classical Probability
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 <i>N</i> 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 \(k, a, b, c \in \mathbb{N}\) with \(a > b > c\). Then:</p>
<p>(a) the value of \(a\) is 52.</p>
<p>(b) the value of \(b+c\) is 52.</p>
<p>(c) the value of \(a+c\) is 52.</p>
<p>(d) 17 is a factor of \(k\).</p>
Step-by-Step Solution
Key Concept: For N=n (getting 2nd ace on nth draw), we need exactly 1 ace in first (n-1) draws and the nth card must be an ace. Use conditional probability: P(N=n) = P(1 ace in first n-1)·P(nth is ace | 1 ace in first n-1).
<p><strong>Step 1: Set up the probability model</strong></p><p>For N=n, we need: exactly 1 ace in first (n-1) draws AND nth draw is an ace.</p><p>P(N=n) = C(4,1)·C(48,n-2)/C(52,n-1) × 3/(52-n+1)</p><p><strong>Step 2: Simplify the expression</strong></p><p>P(N=n) = [4·C(48,n-2)/C(52,n-1)] × 3/(53-n)</p><p>C(48,n-2)/C(52,n-1) = [48!/(n-2)!(50-n)!] / [52!/(n-1)!(53-n)!]</p><p>= (n-1)!(53-n)! / [(n-2)!(50-n)!·52!] = (n-1)(53-n)(52-n)(51-n) / (52·51·50)</p><p><strong>Step 3: Combine terms</strong></p><p>P(N=n) = [4·3(n-1)(53-n)(52-n)(51-n)] / [52·51·50·(53-n)]</p><p>= 12(n-1)(52-n)(51-n) / (52·51·50)</p><p>= (1/1325)(n-1)(52-n)(51-n)</p><p><strong>Step 4: Identify parameters</strong></p><p>P(N=n) = (1/1325)(n-1)(n-52)(n-51) is not in standard form. Rewrite as:</p><p>P(N=n) = (1/1325)(n-1)(-(52-n))(-(51-n)) = (1/1325)(n-1)(52-n)(51-n)</p><p>Comparing with (1/k)(n-a)(n-b)(n-c): k=1325, a=52, b=51, c=1</p><p>Since a > b > c: a=52, b=51, c=1</p><p>∴ Answer: A, B, D</p>
Correct Answer: A,B,D