Question

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

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

Answer 1

The correct option is C $\frac{{}^{20}{C}_{10}}{2^{20}}$

$X$ is a set containing $10$ elements.

Then power set $P\left(X\right)$ contains $2^{10}$ elements.
$A$ and $B$ are picked at random from $P\left(X\right)$ with replacement.
Total no. of outcomes are $2^{10}.$
no. of ways in which $A$ and $B$ have equal no of elements is ${}^{10}{C}_0^{10}{C}_0{+}^{10}{C}_1^{10}{C}_1{+}^{10}{C}_2^{10}{C}_2+....{+}^{10}{C}_{10}^{10}{C}_{10}$
$\therefore$ Probability $=\frac{{}^{10}{C}_0^{10}{C}_0{+}^{10}{C}_1^{10}{C}_1{+}^{10}{C}_2^{10}{C}_2+....{+}^{10}{C}_{10}^{10}{C}_{10}}{2^{20}}$
$=\frac{{}^{20}{C}_{10}}{2^{20}}$