There are $2^n$ tickets in a jar.The frequency of the tickets of number $i$ is ${n \choose i}$ There are $2^n$ tickets in a jar.The frequency of the tickets of number $i$ is ${n \choose i}$ where $i=0,1,2,..n$.
$m$ tickets are drawn randomly without replacement.
Let S be the sum of the numbers drawn.Find $E(S)$ and variance of S.
I can't find a way to approach this problem.
 A: Can you find the expectation of the value of one ticket?  Now use the linearity of expectation.  Can you find the variance of the value of one ticket?  Look up the definition.  What is the variance of a sum?
A: You can easily deduce that, with $X$ the ticket number variable : $$\forall i \in [0,n], P(X = i) = {{n\choose i}\over 2^n}$$
From there, you can apply the formula for expectation :
$$E(X) = \sum_{i=0}^n P(X=i)\cdot i=\sum_{i=0}^n {{n\choose i}\over 2^n}\cdot i$$
Then the formula for variance :
$$V(X) = E[(X-E(X))^2]$$
A: Without resorting to generating functions we can do the following:
Let $X_i$ denote the number on the $i$th ticket for $i=1,2,...,m$, so that $S=\sum\limits_{i=1}^mX_i$.
The pmf of $X_i$ for all $i=1,2,\ldots,m$ is then $$P(X_i=k)=\begin{cases}\frac{1}{2^n}\binom{n}{k}&,\text{ if } k=0,1,\ldots,n\\\\\quad0&,\text{ otherwise}\end{cases}$$
Since $\operatorname{E}(X_i)=\frac{n}{2}$, it follows that $$\operatorname{E}(S)=\frac{mn}{2}$$
Again $\operatorname{E}(X_i^2)=\frac{n(n+1)}{4}$, giving $$\operatorname{Var}(X_i)=\operatorname{E}(X_i^2)-(\operatorname{E}(X_i))^2=\frac{n}{4}$$
The calculation of variance of $S$ is a bit more involved.
\begin{align}
\operatorname{Var}(S)&=\sum_{i=1}^m\operatorname{Var}(X_i)+2\sum_{i<j}\operatorname{Cov}(X_i,X_j)
\\&=m\cdot\frac{n}{4}+2\binom{m}{2}\rho\cdot\frac{n}{4}\qquad,\small\rho\text{ is the correlation  between $X_i$ and $X_j$}
\\&=\frac{mn}{4}\left(1+(m-1)\rho\right)\tag{1}
\end{align}
Now observe that $$\operatorname{Var}\left(\sum_{i=1}^{\color\green{2^n}}X_i\right)=0\tag{2}$$, the sum within parentheses being a constant. 
Moreover the joint distribution of $(X_i,X_j)$ for all $i\ne j$ is independent of $m$. 
So we can replace $m$ by $2^n$ in $(1)$ to get from $(2)$:
$$\frac{n2^n}{4}(1+(2^n-1)\rho)=0$$ 
That is, $$\rho=\frac{1}{1-2^n}$$ 
Substituting the value of $\rho$ in $(1)$, we get $$\operatorname{Var}(S)=\frac{mn}{4}\left(1+\frac{m-1}{1-2^n}\right)$$
