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

Answer 1

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

Total number of subsubsets of set $X=2^{10}=1024$
Number of subsets with zero element ${=}^{10}{C}_0$
Number of subsets with one element ${=}^{10}{C}_1$
Number of subsets with two elements ${=}^{10}{C}_2$
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.
.
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Number of subsets with 10 elements ${=}^{10}{C}_{10}$
$A$ & $B$ are taken from $P\left(X\right)$ from $2^{10}$ subsets so total ways $=2^{10}\times 2^{10}$
Number of ways such that $A$ and $B$ have equal number of elements$={(}^{10}{C}_0{)}^2+{(}^{10}{C}_1{)}^2+{(}^{10}{C}_2{)}^2+........+{(}^{10}{C}_{10}{)}^2$ ${=}^{20}{C}_{10}$
Required probability =$\frac{{}^{20}{C}_{10}}{2^{20}}$