Answer:
E(M) = 23/3
Var(M) = 53/9.
Step-by-step explanation:
There are three possible values for M: 4, 6, and 10.
[tex]\begin{array}{|l|l|}\cline{1-2}\\[-1em]\text{Larger Number} & \text{Smaller Number}\\ \cline{1-2}\\[-1em]4 & 2 \\\cline{1-2}\\[-1em]6 & 2, ~4\\\cline{1-2}\\[-1em] 10 & 2, ~4, ~6\\\cline{1-2} \end{array}[/tex].
That's six unique combinations in total. If the balls are drawn randomly, the probability for getting each combination shall be equal. That is:
[tex]\begin{array}{|c|c|}\cline{1-2}\\[-1em]m & P(\text{M} = m)\\\cline{1-2}\\[-1em] 2 & 0\\\cline{1-2}\\[-1em] 4 & 1/6 \\\cline{1-2}\\[-1em] 6 & 2/6\\\cline{1-2}\\[-1em] 10 & 3/6\\\cline{1-2}\end{array}[/tex].
Consider the formula for the expected value of a discrete random variable:
[tex]\begin{aligned} E({\rm M})& = \sum{[m \cdot P(\mathrm{M} = m)]}\\ &= 4 \times \frac{1}{6} + 6\times \frac{2}{6} + 10 \times \frac{3}{6}\\ &= \frac{23}{3}\end{aligned}[/tex].
Formula for the variance of a discrete random variable (note that this formula can take many other forms):
[tex]\begin{aligned}Var(\mathrm{M}) &= \sum{[m^{2}\cdot P(\mathrm{M}= m)]} - \mu^{2}\\&= 4^{2}\times \frac{1}{6} + 6^{2} \times \frac{2}{6} + 10^{2} \times \frac{3}{6} - \left(\frac{23}{3}\right)^{2}\\&= \frac{53}{9}\end{aligned}[/tex].
