Question

A box contains four bowling balls numbered 2, 4, 6, and 10 according to their weight. Two are drawn randomly without replacement. Let the random variable M denote the max of the two numbers on the balls drawn. Calculate E(M) and Var(M). Hint: Start by specifying the joint distribution of the number on the balls.

Answer 1

E(M) = 23/3

Var(M) = 53/9.

Explanation:

There are three possible values for M: 4, 6, and 10.

`\begin{array}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}`.

That's six unique combinations in total. If the balls are drawn randomly, the probability for getting each combination shall be equal. That is:

`\begin{array}\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}`.

Consider the formula for the expected value of a discrete random variable:

`\begin{aligned} E({\rm M})& = \sum{[m \cdot P(\mathrm{M} = m)]}\\ &= 4 * (1)/(6) + 6* (2)/(6) + 10 * (3)/(6)\\ &= (23)/(3)\end{aligned}`.

Formula for the variance of a discrete random variable (note that this formula can take many other forms):

`\begin{aligned}Var(\mathrm{M}) &= \sum{[m^(2)\cdot P(\mathrm{M}= m)]} - \mu^(2)\\&= 4^(2)* (1)/(6) + 6^(2) * (2)/(6) + 10^(2) * (3)/(6) - \left((23)/(3)\right)^(2)\\&= (53)/(9)\end{aligned}`.