Question

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 :

• A. $1/2$
• B. $1/\left(2^k-1\right)$
• C. $1/2^k$
• D. $1/3^k$

Answer 1

Answer: (B)

$n(X)=k$, Number of subsets of $X=2^k$
The number of elements and the number of sets are given by the binomial expansion
$2^k={}^k{C}_0+{}^k{C}_1+{}^k{C}_2+\dots .+{}^k{C}_{k-2}+{}^k{C}_{k-1}+{}^k{C}_k$
${}^k{C}_0=$ 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, $P(E)=\frac{m\cdot 2}{2\cdot m(m-1)}=\frac{1}{m-1}=\frac{1}{\left(2^k-1\right)}$