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Let $X$ be a set containing $n$ elements. If two subsets $A$ and $B$ of $X$ are picked at random, the probability that $A$ and $B$ have the same number of elements, is
$\frac{{^{2n}{C_n}}}{{{2^{2n}}}}$
$\frac{1}{{^{2n}{C_n}}}$
$\frac{{1\,.\,3\,.\,5......(2n - 1)}}{{{2^n}}}$
$\frac{{{3^n}}}{{{4^n}}}$
Solution
(a) We know that the number of sub-sets of a set containing $n$ elements is ${2^n}.$
Therefore the number of ways of choosing $A$ and $B$ is ${2^n}.$ ${2^n} = {2^{2n}}$
We also know that the number of sub-sets (of $X$) which contain exactly $r$ elements is ${}^n{C_r}.$
Therefore the number of ways of choosing $A$ and $B,$ so that they have the same number elements is
${({}^n{C_0})^2} + {({}^n{C_1})^2} + {({}^n{C_2})^2} + \,…… + {({}^n{C_n})^2} = {}^{2n}{C_n}$
Thus the required probability $ = \frac{{{}^{2n}{C_n}}}{{{2^{2n}}}}.$
