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

Answer 1

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

Total powersets of $x=2^{10}$

No. of way equal elements power sets are chosen

${}^{10}{C}_1{\times }^{10}{C}_1{+}^{10}{C}_2{\times }^{10}{C}_2+.....{.}^{10}{C}_{10}{\times }^{10}{C}_{10}$

$=\sum _{i=1}^{10}{(}^{10}{C}_1{)}^2$

${=}^{20}{C}_{10}$

probability $=\frac{{}^{20}{C}_{10}}{2^{10}\times 2^{10}}=\frac{{}^{20}{C}_{10}}{2^{20}}$

$D$ is correct