Expectation and variance of the number of elements of a random non-empty set selected from a finite power set Let $S$ denote a finite set of cardinality $|S| = N$. Select randomly a non-empty subset of $S$. Let $X$ indicate the number of items belonging to this subset.
(a) Describe the probability mass function of $X$ and compute $\textbf{E}(X)$ and $\textbf{Var}(X)$.
(b) Prove that $\displaystyle\lim_{N\rightarrow\infty}\frac{\textbf{E}(X)}{N} = \frac{1}{2}$ and $\displaystyle\lim_{N\rightarrow\infty}\frac{\textbf{V}(X)}{N} = \frac{1}{4}$.

MY ATTEMPT
(a) In the first place, we first notice there are $C(N,x)$ subsets which has $x$ elements, where $1\leq x\leq N$. Therefore we conclude that
\begin{align*}
f_{X}(x) = \frac{1}{2^{N} - 1}{N\choose x}
\end{align*}
From then on we are able to determine $\textbf{E}(X)$ and $\textbf{Var}(X)$. Precisely speaking, we have
\begin{align*}
\sum_{x=1}^{N}{N\choose x}x &amp; = \sum_{x=1}^{N}\frac{N!}{(x-1)!(N-x)!} = \sum_{x=0}^{N-1}\frac{N!}{x!(N-x-1)!}\\
&amp; = N\sum_{x=0}^{N-1}\frac{(N-1)!}{x!(N-x-1)!} = N\sum_{x=0}^{N-1}{N-1\choose x} = N2^{N-1}
\end{align*}
according to the binomial theorem. From whence we conclude that
$$\textbf{E}(X) = \frac{N2^{N-1}}{2^{N}-1}$$
Further, we do also have the following identity
\begin{align*}
\sum_{x=1}^{N}{N\choose x}x^{2} &amp; = \sum_{x=1}^{N}\left[\frac{N!}{(x-1)!(N-x)!}\right]x = \sum_{x=0}^{N-1}\left[\frac{N!}{x!(N-x-1)!}\right](x+1)\\
&amp; = \sum_{x=1}^{N-1}\frac{N!}{(x-1)!(N-x-1)!} + \sum_{x=0}^{N-1}\frac{N!}{x!(N-x-1)!}\\
&amp; = N(N-1)\sum_{x=0}^{N-2}\frac{(N-2)!}{x!(N-x-2)!} + N\sum_{x=0}^{N-1}\frac{(N-1)!}{x!(N-x-1)!}\\
&amp; = N(N-1)\sum_{x=0}^{N-2}{N\choose x} + N\sum_{x=0}^{N-1}{N-1\choose x} = N(N-1)2^{N-2} + N2^{N-1}
\end{align*}
Therefore
$$\textbf{E}(X^{2}) = \frac{N(N-1)2^{N-2} + N2^{N-1}}{2^{N}-1}$$
Finally, we have the variance:
\begin{align*}
\frac{\textbf{Var}(X)}{N} &amp; = \frac{\textbf{E}(X^{2}) - \textbf{E}(X)^{2}}{N}
\end{align*}
(b) As requested, we obtain the desired result
\begin{align*}
\lim_{N\rightarrow\infty}\frac{\textbf{E}(X)}{N} = \lim_{N\rightarrow\infty}\frac{2^{N-1}}{2^{N}-1} = \frac{1}{2}
\end{align*}
However, the same does not apply to the variance when I make use of the results obtained. Am I committing some conceptual mistake? I would be equally grateful if someone provided me a quicker or smarter way to tackle this problem.
EDIT
I edited my answer as suggested by Did. Nonetheless, I am still thinking there is something wrong with my solution. Could someone double-check my calculations? Thanks in advance.
Seem probably simpler if you look at it this way: Say $S$ has $n$ elements. If  $Y$ is the number of elements of a randomly selected element of the power set, including the empty set as a possibility, then $Y$ has the same distribution as $X_1+\dots+X_n$, where the $X_j$ are iid with $P(X_1=0)=P(X_1=1)=1/2$.
Seem probably simpler if you look at it this way: Say $S$ has $n$ elements. If  $Y$ is the number of elements of a randomly selected element of the power set, including the empty set as a possibility, then $Y$ has the same distribution as $X_1+\dots+X_n$, where the $X_j$ are iid with $P(X_1=0)=P(X_1=1)=1/2$.