Question:
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:
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:
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Count subsets of each size k with \(^{10}C_k\) and use \(\sum {}^{10}C_k^2={}^{20}C_{10}\).
Updated On: Nov 11, 2026
- \(\frac{2^{10}-1}{2^{20}}\)
- \(\frac{{}^{20}C_{10}}{2^{20}}\)
- \(\frac{{}^{20}C_{10}}{2^{10}}\)
- \(\frac{2^{10}-1}{2^{10}}\)
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The Correct Option is B
Solution and Explanation
P(X) has \(2^{10}\) elements, so the total number of ordered choices (A, B) with replacement is \(2^{10}\cdot 2^{10}=2^{20}\).
For a given size k there are \({}^{10}C_k\) subsets, so favourable pairs with |A| = |B| = k number \(({}^{10}C_k)^2\).
Favourable total \(=\sum_{k=0}^{10}({}^{10}C_k)^2={}^{20}C_{10}\) (Vandermonde identity).
Probability \(=\frac{{}^{20}C_{10}}{2^{20}}\).
Answer: (2) \(\frac{{}^{20}C_{10}}{2^{20}}\)
For a given size k there are \({}^{10}C_k\) subsets, so favourable pairs with |A| = |B| = k number \(({}^{10}C_k)^2\).
Favourable total \(=\sum_{k=0}^{10}({}^{10}C_k)^2={}^{20}C_{10}\) (Vandermonde identity).
Probability \(=\frac{{}^{20}C_{10}}{2^{20}}\).
Answer: (2) \(\frac{{}^{20}C_{10}}{2^{20}}\)
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