Question

A pair of fair dice is thrown. Let X be the random variable which denotes the minimum of the two numbers which appear. Find the probability distribution, mean and variance of X.

Answer 1

Let X denote the event of getting twice the number. Then, X can take the values 1, 2, 3, 4, 5 and 6.

Thus, the probability distribution of X is given by

X	P(X)
1	$\frac{11}{36}$
2	$\frac{9}{36}$
3	$\frac{7}{36}$
4	$\frac{5}{36}$
5	$\frac{3}{36}$
6	$\frac{1}{36}$

Computation of mean and variance

xi	pi	pixi	pixi2
1	$\frac{11}{36}$	$\frac{11}{36}$	$\frac{11}{36}$
2	$\frac{9}{36}$	$\frac{18}{36}$	1
3	$\frac{7}{36}$	$\frac{21}{36}$	$\frac{63}{36}$
4	$\frac{5}{36}$	$\frac{20}{36}$	$\frac{80}{36}$
5	$\frac{3}{36}$	$\frac{15}{36}$	$\frac{75}{36}$
6	$\frac{1}{36}$	$\frac{6}{36}$	1
		${\sum }_{}^{}$pixi = $\frac{91}{36}=2.5$	${\sum }_{}^{}$pixi2 = $\frac{301}{36}=8.4$

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=2.5 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2=8.4-6.25=2.15 \end{gathered}$$