Problem source

In a bag are $n$ balls numbered 1, 2, $\ldots\,$, $n\,$. When a ball is taken out of the bag, each ball is equally likely to be taken. 
\begin{questionparts}
\item A ball is taken out of the bag. The number on the ball is noted and the ball is replaced in the bag. The process is repeated once.
Explain why the expected value of the product of the numbers on the two balls is 
\[
\frac 1 {n^2}  \sum_{r=1}^n\sum_{s=1}^n rs 
\]
and simplify this expression.
\item
 A ball is taken out of the bag. The number on the ball is noted and the ball is \textit{not} replaced in the bag. Another ball is taken out of the bag and the number on this ball is noted.
Show that the expected value of the product of the two numbers is 
\[
\frac{(n+1)(3n+2)}{12}\;.
\]
\end{questionparts}
\textbf{Note: } $\displaystyle \sum_{r=1}^n r = \frac12 n(n+1)$ and $\displaystyle \sum_{r=1}^n r^2 = \frac16 n(n+1)(2n+1)\;$.

Solution source

\begin{questionparts}
\item Since the second draw is independently of the first draw $\mathbb{P}(F=r, S=s) = \mathbb{P}(F=r)\mathbb{P}(S=s) = \frac{1}{n^2}$ and so
\begin{align*}
&& \E[FS] &= \sum_{r=1}^n \sum_{s=1}^n rs \mathbb{P}(F=r, S = s) \\
&&&= \frac{1}{n^2} \sum_{r=1}^n \sum_{s=1}^n rs \\
&&&= \frac{1}{n^2} \left ( \sum_{r=1}^n r \right)\left ( \sum_{s=1}^n s \right) \\
&&&= \frac{1}{n^2} \left ( \frac{n(n+1)}{2} \right)^2 \\
&&&= \frac{(n+1)^2}{4}
\end{align*}

\item Under this methodology, $\mathbb{P}(F=r, S=s) = \frac{1}{n(n-1)}$ as long as $r \neq s$, therefore
\begin{align*}
&& \E[FS] &= \sum_{r=1}^n \sum_{s\neq r} rs \mathbb{P}(F = r, S = s) \\
&&&= \frac{1}{n(n-1)} \sum_{r=1}^n \left ( \sum_{s=1}^n rs - r^2 \right) \\
&&&= \frac{1}{n(n-1)}  \left ( \frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6} \right) \\
&&&= \frac{n+1}{12(n-1)}  \left ( 3n(n+1)-2(2n+1) \right) \\
&&&= \frac{n+1}{12(n-1)} \left ( 3n^2 -n -2 \right) \\
&&&= \frac{n+1}{12(n-1)} (3n+2)(n-1)\\
&&&= \frac{(n+1)(3n+2)}{12}
\end{align*}
\end{questionparts}