<p>Let \(X\) be a set containing 10 elements and \(P(X)\) be its power set. If \(A\) and \(B\) are picked up at random from \(P(X)\), with replacement, then the probability that \(A\) and \(B\) have equal number of elements, is</p>
<p>\(\dfrac{{}^{20}C_{10}}{2^{10}}\)</p>
<p>\(\dfrac{(2^{10}-1)}{2^{20}}\)</p>
<p>\(\dfrac{(2^{10}-1)}{2^{10}}\)</p>
<p>\(\dfrac{{}^{20}C_{10}}{2^{20}}\)</p>
Step-by-Step Solution
Key Concept: The probability that two randomly selected subsets have equal cardinality equals the sum of probabilities for each possible cardinality k, where each probability is [C(n,k)/2^n]². This uses the fact that the number of k-element subsets is C(n,k) and total subsets is 2^n.
<p><strong>Step 1:</strong> Set X has 10 elements, so |P(X)| = 2^10 = 1024 subsets.</p><p><strong>Step 2:</strong> Total ways to pick (A, B) with replacement = (2^10)² = 2^20.</p><p><strong>Step 3:</strong> For A and B to have equal number of elements, both must have exactly k elements for some k ∈ {0, 1, 2, ..., 10}.</p><p><strong>Step 4:</strong> Number of subsets with exactly k elements = C(10, k).</p><p><strong>Step 5:</strong> For a fixed k, favorable outcomes = C(10, k) × C(10, k) = [C(10, k)]²</p><p><strong>Step 6:</strong> Total favorable outcomes = Σ(k=0 to 10) [C(10, k)]² = C(20, 10) (by Vandermonde's identity)</p><p><strong>Step 7:</strong> Probability = C(20, 10) / 2^20 = 184756 / 1048576 ≈ 0.176</p><p>∴ Answer: <strong>C(20, 10) / 2^20</strong> or approximately <strong>D</strong></p>
Correct Answer: D