Question

Let $U$ be the universal set and $n\left(x\right)=k+1$, the probability
of selecting $2$ subsets $A$ and $B$ of the set $X$ such that $B=A^c$ is

• A. $\frac{1}{2^k-1}$
• B. $\frac{2}{2^k-1}$
• C. $\frac{1}{2\cdot 2^k-1}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2\cdot 2^k-1}$

$n\left(X\right)=K+1$
$\therefore$ Number of subsets of $X$ are $2^{k+1}$.
$\therefore$ Number of sample points of $S$ are ${}^{2^{k+1}}{\text{C}}_2$
(As $A$ and $B$ are subset of $X$ to be selected)
Now $n\left(A\right)=$ Number of selections of two non-intersecting subset whose union is $X$.
$=\frac{1}{2}[{}^{2^{k+1}}{\text{C}}_0{+}^{2^{k+1}}{\text{C}}_1+\cdots ]$
($\because$ the number of selections in which each one subset has I elements and the rest are in the other subset ${}^{k+1}{C}_1$, and every selection occurs twice in all (total of) the selections.
$\therefore P\left(A\right)=\frac{1}{2}\frac{\frac{2^{k+1}}{2^{k+1}(2^{k+1}-1)}}{2}$
$\because {}^{2^{k+1}}{\text{C}}_2=\frac{2^{k+1}}{2}\cdot \frac{2^{k+1}-1}{1}$
$=\frac{1}{2}\frac{2^{k+1}}{2^{k+1}}\times \frac{2}{2^{k+1}-1}$ or ${}^n{\text{C}}_r=\frac{n}{r}{\cdot }^{n-1}{\text{C}}_{r-1}$ $=\frac{1}{2\times 2^k-1}$ or $\frac{1}{2^{k+1}-1}$