<p>From an ordinary pack of 52 cards an even number of cards are drawn at random. Find the probability of getting equal number of black and red cards.</p>
<p>\(\dfrac{^{52}C_{26}}{2^{52}-1}\)</p>
<p>\(\dfrac{^{52}C_{26}-1}{2^{51}-1}\)</p>
<p>\(\dfrac{^{52}C_{26}}{2^{51}-1}\)</p>
<p>\(\dfrac{^{52}C_{26}+1}{2^{52}-1}\)</p>
Step-by-Step Solution
Key Concept: The total number of ways to draw an even number of cards equals 2^51 (by symmetry of binomial coefficients). For equal black-red cards, we need to sum C(26,k)² over all valid k, which also equals 2^51 by Vandermonde's identity.
<p><strong>Step 1:</strong> Identify the sample space. Drawing an even number of cards from 52 means drawing 2, 4, 6, ..., or 52 cards. Total ways = C(52,2) + C(52,4) + ... + C(52,52) = 2^51 (by binomial theorem applied to (1+1)^52 and separating even-indexed terms).</p><p><strong>Step 2:</strong> Count favorable outcomes (equal black and red). For drawing 2k cards with k black and k red: we need Σ C(26,k) · C(26,k) for k = 1, 2, ..., 26.</p><p><strong>Step 3:</strong> Apply Vandermonde's identity. By the identity Σ C(m,k)·C(n,k) = C(m+n,m), we have Σ C(26,k)² = C(52,26). However, the sum of favorable outcomes (considering all even draws with equal colors) equals 2^51 by the symmetry argument: exactly half of all even-card draws have equal black-red.</p><p><strong>Step 4:</strong> Calculate probability. P(equal black and red | even draw) = 2^51/2^51 = <strong>1/2</strong></p><p>∴ Answer: B</p>
Correct Answer: B