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Q.
Let $X$ be a universal set such that $n(X)=k$. The probability of selecting two subsets $A$ and $B$ of the set $X$ such that $B=\overline{A}$ is :
Probability
Solution:
$n(X)=k$, Number of subsets of $X=2^{k}$
The number of elements and the number of sets are given by the binomial expansion
$2^{k}={ }^{k} C_{0}+{ }^{k} C_{1}+{ }^{k} C_{2}+\ldots .+{ }^{k} C_{k-2}+{ }^{k} C_{k-1}+{ }^{k} C_{k}$
${ }^{k} C_{0}=$ null set (without any element) and ${ }^{k} C_{k}$ (universal set) are complementary.
Similarly there are $k$ singletons $\left({ }^{k} C_{1}\right)$ which will have $k$ sets with $(k-1)$ elements each as their complementary sets.
$\therefore $ No. of such combinations $=1 / 2\left(2^{k}\right)$
Now two subsets from $2^{k}$ subsets can be selected in
${ }^{2^{k}} C_{2}$ ways
$\therefore $ Required probability $P(E)=\frac{2^{k}}{2 \cdot{ }^{2^{k}} C_{2}}$
Put $2^{k}=m$
So, $P(E)=\frac{m \cdot 2}{2 \cdot m(m-1)}=\frac{1}{m-1}=\frac{1}{\left(2^{k}-1\right)}$