Question

Let $x$ be a set containing $n$elements. If two subsets$A$ and $B$of $x$picked at random, the probability that$A$ and $B$have the same number of elements, is

• A. $\frac{{}^{2n}{\mathrm{ℂ}}_n}{2^{2n}}$
• B. $\frac{1}{{}^{2n}{\mathrm{ℂ}}_n}$
• C. $\frac{\left[1.3.5...\left(2n-1\right)\right]}{2^n}$
• D. $\frac{3^n}{4^n}$

Answer 1

The correct option is A

$\frac{{}^{2n}{\mathrm{ℂ}}_n}{2^{2n}}$

Explanation for the correct option:

Determine the probability of $A$ and $B$being the same number.

The number of subsets in a set contains $n$ elements.is $2^n$

Hence, the total number of ways to choose $A\text{and}B$ is $2^n$.$2^n$or $2^{2n}$

Similarly, the number of subsets of $A$, that contains exactly$r$elements is ${}^n{\mathrm{ℂ}}_r$

Hence, the favorable cases is $\left({}^n{\mathrm{ℂ}}_0\right)\left({}^n{\mathrm{ℂ}}_0\right)+\left({}^n{\mathrm{ℂ}}_1\right)\left({}^n{\mathrm{ℂ}}_1\right)+...+\left({}^n{\mathrm{ℂ}}_n\right)\left({}^n{\mathrm{ℂ}}_n\right)$

$$\begin{gathered} =\left({}^n{\mathrm{ℂ}}_0\right)\left({}^n{\mathrm{ℂ}}_{n-0}\right)+\left({}^n{\mathrm{ℂ}}_1\right)\left({}^n{\mathrm{ℂ}}_{n-1}\right)+...+\left({}^n{\mathrm{ℂ}}_n\right)\left({}^n{\mathrm{ℂ}}_{n-n}\right) \\ {=}^{2n}{\mathrm{ℂ}}_n \end{gathered}$$

Therefore, the required probability is $\frac{{}^{2n}{\mathrm{ℂ}}_n}{2^{2n}}$

Hence, option (A) is the correct answer.