Five Marbles

The number of discrete outcomes $N = \displaystyle {5 \choose 2} = \dfrac {5 \times 4}{2 \times 1} = 10$ . Therefore, the probability for an outcome $p = \frac 1{10}$ . Therefore the expected value of the outcome is:

$\begin{aligned} E(X) & = \sum_{i < j}^5 ij p = \frac 1{10} \sum_{i=1}^4 \sum_{j=i+1}^5 ij = \frac 1{10} \sum_{i=1}^4 i \sum_{j=i+1}^5 j \\ & = \frac 1{10} \sum_{i=1}^4 \frac {i(5-i)(i+6)}2 \\ & = \frac 1{20} \sum_{i=1}^4 \left(30i-i^2-i^3\right) \\ & = \frac {30(10)-30-100}{20} \\ & = \boxed{8.5} \end{aligned}$

There are 5 choose 2 = 10 ways to pick 2 marbles from 5 marbles.

The expected value of the product is: 1/10 x (1x2+1x3+1x4+1x5+2x3+2x4+2x5+3x4+3x5+4x5) = 1/10 x 85 = 8.5