What is the average product of the numbers drawn? There are 10 tickets in a box, each with a number on it. The average of those 10 numbers is zero. The average value of the squares of those numbers is 5.
(a). If you pick two tickets at random, WITH replacement, what is the average product of the numbers drawn?
(b). If you pick two tickets at random, WITHOUT replacement, what is the average product of the numbers drawn?
 A: Since there's now in my view an overly formal and an insufficiently formal answer, I'll offer what I see as the appropriate middle ground :-)
Let $a_k$ denote the $10$ numbers.
(a)
$$\sum_i\sum_j(a_ia_j)=\left(\sum_ia_i\right)\left(\sum_ja_j\right)=0\;.$$
The sum of all products is zero, so the average product is zero.
(b)
$$\sum_i\sum_{j\ne i}(a_ia_j)=\sum_i\sum_j(a_ia_j)-\sum_i(a_ia_i)=0-10\cdot5=-50\;.$$
The sum contains $10\cdot9=90$ products, so the average product is $\frac{-50}{90}=-\frac59$.
A: The average sum is zero and average value of squares is $5.$
Let the numbers be $-\sqrt{5}\mbox{  }$ $\mbox{  }5$ times and $\sqrt{5}$ $f$ times.
$a)$ Let first ticket be $\sqrt{5}$ and the second ticket can be $\pm$ with equal probability.  Average product $=0$
$b)$ If first number is $\sqrt{5}$. Second number can be positive with $\dfrac49$ probability and negative with $\dfrac59$ probability . Average product $=-5\times\dfrac19=-\dfrac59$
Similar case if the first number is $-\sqrt{5}$ and Average $=-\dfrac59$
So, the average product $=-\dfrac59$
A: Let $n=10$, this makes it simpler to type that index in LaTeX. (We cannot speak about a generalization...)
Let $\Omega_0$ be $\{1,2,\dots n\}$, we consider on it the uniform probability. 
The question deals with $\Omega =\Omega_0\times\Omega_0$, the set of all $(j,k)$, the case "with replacement", and with 
$\Omega'=\{\ (j,k)\in\Omega\ :\ j\ne k\ \}$, the case "without replacement".
We consider the random variable $A:\Omega_0\to \Bbb R$, $j\to A(j)=a_j$, and the induced variables $X:\Omega\to\Bbb R$, $X(j,k)=A(j)A(k)$ on $\Omega$, respectively $X' :\Omega'\to\Bbb R$, given by the same formula $X'(j,k)=A(j)A(k)$, but the domain of definition is smaller.
The problem gives:
$$
\begin{aligned}
\Bbb E[A] &=\frac 1n\sum_{1\le j\le n }a_j=0\ ,\\
\operatorname{Var}[A] &=\frac 1n\sum_{1\le j\le n }(a_j-0)^2=5\ .
\end{aligned}
$$
(So $\sum a_j^2=5n$ to have it simpler.)
Let us start to solve the problem.
$$
\begin{aligned}
\Bbb E[X] 
&=\frac 1{|\Omega|}\sum_{(j,k)\in\Omega }X(j,k)
=\frac 1{n^2}\sum_{1\le j,k\le n }a_ja_k
=\frac 1{n^2}
\left(\sum_{1\le j\le n }a_j\right)
\left(\sum_{1\le k\le n }a_k\right)
\\
&=(\ \Bbb E A\ )^2 =0^2 =0\ ,
\\[2mm]
\Bbb E[X'] 
&=\frac 1{|\Omega'|}\sum_{(j,k)\in\Omega }X'(j,k)
=\frac 1{n(n-1)}\sum_{\substack{1\le j,k\le n\\j\ne k}}a_ja_k\\
&=\frac 1{n(n-1)}
\sum_{1\le j,k\le n}a_ja_k 
-
\frac 1{n(n-1)}
 \sum_{\substack{1\le j,k\le n\\j=k}}a_ja_k 
\\
&=
\frac 1{n(n-1)}
\left[ \ 
\left(\sum_{1\le j\le n }a_j\right)
\left(\sum_{1\le k\le n }a_k\right)
-
\left(\sum_{1\le j\le n }a_j^2\right)
\ \right]
\\
&=
\frac n{n-1}(\Bbb E A)^2
-
\frac 1{n-1}\operatorname{Var}[A]
\\
&=
-
\frac 1{n-1}\operatorname{Var}[A]
=-\frac 59\ .
\end{aligned}
$$
(Sorry about the "probability space bureaucracy"... Somehow i felt the obligation to do it.)
