Question

Let $X$ be a set containing $10$ elements and $P(X)$ be its power set. If $A$ and $B$ are picked up at random from $P(X)$, with replacement, then the probability that $A$ and $B$ have equal number of elements, is :

• A. $\frac{{}^{20}{C}_{10}}{2^{10}}$
• B. $\frac{\left(2^{10}-1\right)}{2^{20}}$
• C. $\frac{\left(2^{10}-1\right)}{2^{10}}$
• D. $\frac{{}^{20}{C}_{10}}{2^{20}}$

Answer 1

Answer: (D)

$n(x)=10,n(P(x))=(2{)}^{10}=1024$
$A,B\in P(x)$
$n$ (sample space) $=2^{10}\times 2^{10}$
$n(E)=\left(1\times 1+{}^{10}{C}_1\times {}^{10}{C}_1+{}^{10}{C}_2\times {}^{10}{C}_2+{}^{10}{C}_3\times {}^{10}{C}_3+\dots +{}^{10}{C}_{10}\times {}^{10}{C}_{10}\right)$
$={\left({}^{10}{C}_0\right)}^2+{\left({}^{10}{C}_1\right)}^2+{\left({}^{10}{C}_2\right)}^2+\dots +{\left({}^{10}{C}_{10}\right)}^2$
$={}^{20}{C}_{10}$
$\therefore P(E)=\frac{{}^{20}{C}_{10}}{2^{10}\times 2^{10}}=\frac{{}^{20}{C}_{10}}{2^{20}}$