<p>A bag contains \(n\) white and \(n\) black balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each pair consists of one white and one black ball is</p>
Step-by-Step Solution
Key Concept: Each pair must contain one white and one black ball, which means after drawing 2n balls in n pairs, we need exactly n white and n black balls arranged alternately in draw order. Use the sequential probability approach: at each draw step, calculate the probability of getting the required color.
<p><strong>Step 1:</strong> Set up the sequential drawing process. We draw 2n balls in n pairs. For each pair to have exactly 1 white and 1 black ball, at positions (1,2), (3,4), (5,6), ..., (2n-1, 2n), we need alternating colors.</p><p><strong>Step 2:</strong> Calculate the probability of one valid arrangement (say W-B-W-B-...-W-B). First draw white: n/(2n), then black: n/(2n-1), then white: (n-1)/(2n-2), then black: (n-1)/(2n-3), etc.</p><p>P(W-B-W-B-...W-B) = [n·n·(n-1)·(n-1)·...·1·1] / [2n·(2n-1)·(2n-2)·...·2·1]</p><p>= [n!·n!] / [2n!]</p><p><strong>Step 3:</strong> Account for all valid arrangements. We can also start with B-W-B-W-...-B-W (one other valid pattern with same probability). However, the general formula for exactly n pairs of (W,B) requires counting arrangements.</p><p><strong>Step 4:</strong> The probability is:</p><p>P = [n!·n!] / [2n!] = 1/C(2n,n)</p><p>∴ Answer: A</p>
Correct Answer: A