What is the expected value of the product of randomly selected balls 
An urn contains four balls numbered 1, 2, 5, and 7.  If a person
  selects a set of two balls at random, what is the expected value of
  the product of the numbers on the balls?

My thoughts:
$E(X)=\sum_{k=1}^{n}a_{k}b_{k}$
$E(X) = E_{1}(X)*E_{2}(X)$
The probability of choosing the first ball is $\frac{1}{4}$, so the expected value of the first ball is $E_{1}(X)=\tfrac{1}{4}\sum(1+2+5+7) = 3.75$
The probability of choosing the second ball is  $\frac{1}{3}$(because 3 balls left after the first move). But what would be the $E_{2}(X)$, since we don't know which ball was chosen first and can't say $E_{2}(X)=\tfrac{1}{4}\sum(2+5+7)$
 A: There's only 6 different ways to draw two balls, and each of those are equally likely, so I would just calculate it as:
$$
\frac{1\cdot 2+1\cdot 5+1\cdot 7+2\cdot 5+2\cdot 7+5\cdot 7}{6}
=\frac{2+5+7+10+14+35}{6}
=\frac{73}{6}
=12\frac{1}{6}
$$
A: I don't know what you mean by $E_1$ and $E_2$ but it is not true that
$$E[X] = E[X_1]E[X_2]$$
where $X_i$ is the value on the $i$th draw. 
We can approach explicitly for clarity's sake (not efficiency):
\begin{array}{r|c|c|c|c}
 &1&2&5&7\\\hline
1&\mathsf X&2&5&7\\\hline
2&2&\mathsf X&10&14\\\hline
5&5&10&\mathsf X&35\\\hline
7&7&14&35&\mathsf X
\end{array}
Each cell that is not $\mathsf X$ has chance $(1/4)(1/3) = 1/12$.
Notice that the possible values of $X$ are $\mathscr K=\{2,5,7, 10,14,35\}$.
This implies the distribution of $X$ is
\begin{array}{c|c}
k&P(X=k)\\\hline
2&2/12\\\hline
5&2/12\\\hline
7&2/12\\\hline
10&2/12\\\hline
14&2/12\\\hline
35&2/12\\
\end{array}
Hence
$$E[X] = \sum_{k\in \mathscr K} kP(X = k) = \frac{73}{6}.$$
A: The expectation of a sum of random variables is always equal to the sum of expectations of the variables.   That is due to the Linearity of Expectation.
However, the expectation of a product of random variables is not always equal to the product of the expectation unless they are uncorrelated.   These random variables are not uncorrelated.
Instead you can use the tower property, also known as the Law of Iterated Expectation.
$$\begin{align}\mathsf E(X_1{\cdot}X_2)=&~\mathsf E(X_1{\cdot}\mathsf E(X_2\mid X_1))\\[1ex]=&~ \mathsf E(X_1{\cdot}\tfrac13(15-X_1)) \\[1ex]=&~ 5\,\mathsf E(X_1)-\tfrac 13\,\mathsf E(X_1^2) \\[1ex]=&~ \tfrac{75}{4}-\tfrac 1{12}(1+4+25+49)\\[1ex]=&~ \tfrac{73}{6}\end{align}$$
