Question

Let $S$ be the universal set and $n\left(X\right)=k$. The probability of selecting two subsets $A$ and $B$ of the set $X$ such that $B=\overline{A}$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{2^k-1}$
• C. $\frac{1}{2^k}$
• D. $\frac{1}{3^k}$

Answer 1

The correct option is C $\frac{1}{2^k}$

Since $B=\overline{A},A\cup B=X$.
The set $X$ may be divided into two non-overlapping parts $A$ and $B$ where $A$ has $r$ elements and $B$ has the remaining $(k-r)$ elements, where $r=0,1,2,3,...k$.
This can thus be achieved in ${}_0^{k}C{+}_1^{k}C{+}_2^{k}C+...{+}_k^{k}C=2^k$ ways.
Total number of ways of selecting two subsets of $X$ is $2^k\ast 2^k=2^{(k+1)}$
Therefore, the required probability = $\frac{2^k}{2^{(k+k)}}=\frac{1}{2^k}$.