Assume that $1\le k \le n$ and let $\mathscr{Z}$ be the family of all subsets of $\{1,\ldots,n\}$ with at most $k$ elements. Pick a random element $X$ of $\mathscr{Z}$ (we consider the probablity distribution on $\mathscr{Z}$ is uniform, that is, each $X$ is chosen with probability $1/|\mathscr{Z}|$). What is the expected value for the random variable $\xi\_n^k=|X|$, where $|X|$ stands for the cardinality of $X$?


In addition, what can be said about asymptotic behaviour of $\xi\_n^{\lfloor n\delta\rfloor}$ for a fixed $\delta>0$?



Expected cardinality of a randomly chosen element of the family of subsets of $\{1,\ldots,n\}$ with at most $k$-elements
$\DeclareMathOperator\E{E}$As already noted by the other answers,
$$\E\xi^k\_n=\frac{\sum\_{i=0}^ki\binom ni}{\sum\_{i=0}^k\binom ni}.$$
One can then easily determine the asymptotics of $\E\xi\_n^{\lfloor n\delta\rfloor}$ for fixed $0\le\delta\le1$:


**Case 1: $0\le\delta<1/2$.** Then
$$k-\frac{\delta}{1-2\delta}\le\E\xi\_n^k\le k,$$
where $k=\lfloor n\delta\rfloor$. This can be shown by approximation of $\binom ni$ by a geometric series. That is, we have
$$0<i\le k\implies\binom n{i-1}=\frac i{n-i+1}\binom ni\le\frac i{n-i}\binom ni\le\frac\delta{1-\delta}\binom ni,$$
hence 
$$0\le j\le i\le k\implies \binom nj\le\left(\frac\delta{1-\delta}\right)^{i-j}\binom ni.$$


Thus,
$$\begin{align}
k-\E\xi^k\_n=\frac{\sum\_{i=0}^k(k-i)\binom ni}{\sum\_{i=0}^k\binom ni}
&=\frac{\sum\_{j=1}^k\sum\_{i=0}^{k-j}\binom ni}{\sum\_{i=0}^k\binom ni}\\
&\le\frac{\sum\_{j=1}^k\left(\frac\delta{1-\delta}\right)^j\sum\_{i=j}^k\binom ni}{\sum\_{i=0}^k\binom ni}\\
&\le\sum\_{j=1}^k\left(\frac\delta{1-\delta}\right)^j\le\frac\delta{1-2\delta}.
\end{align}$$


One can show that the lower bound is closer to the truth: $\E\xi^k\_n=k-\frac\delta{1-2\delta}+O\bigl(\frac{\log n}n\bigr)$.


**Case 2: $1/2<\delta\le1$.** Then $\E\xi^k\_n=\frac n2-O(\gamma^n)$ for some $\gamma<1$ (depending on $\delta$).


Indeed, Stirling bounds give
$$2^n-\sum\_{i=0}^k\binom ni=\sum\_{i=k+1}^n\binom ni=O(\alpha^n)$$
for some constant $\alpha<2$, hence
$$\E\xi^k\_n=\frac{\sum\_{i=0}^ni\binom ni+O(n\alpha^n)}{\sum\_{i=0}^n\binom ni+O(\alpha^n)}=\frac n2+O\bigl(n(\alpha/2)^n\bigr).$$


**Case 3: $\delta=1/2$.** Then
$$\frac n2-\frac{\sqrt n}2\le\E\xi^k\_n\le\frac n2.$$
Indeed, if $Y$ is drawn from the binomial distribution $B(n,1/2)$, we have
$$\frac n2-\E\xi^k\_n=\E\left|Y-\frac n2\right|\le\sqrt{\E\left(Y-\frac n2\right)^2}=\sqrt{\operatorname{Var}Y}=\frac{\sqrt n}2.$$


In this case, approximation of the binomial distribution by Gaussian distribution with mean $n/2$ and variance $n/4$ suggests that the true value of $\E\xi^k\_n$ should be roughly $\frac n2-\sqrt{\frac{n}{2\pi}}$, but I will not try to make this rigorous.


