Question

Two numbers are selected at random (without replacement) from first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find the probability distribution of $X$. Find the mean and variance of this distribution.

Answer 1

$S=\{1,2,3,4,5,6\}$

Clearly $X$ can take values $2,3,4,5,6$

$P(X=2)$ = getting $1$ in first selection and $2$ in second selection or getting $2$ in first selection and $1$ in second selection.

= $\frac{1}{6}\times \frac{1}{5}+\frac{1}{6}\times \frac{1}{5}=\frac{2}{30}$

$P(X=3)$ = getting less than $3$ in first and $3$ in second selection or getting $3$ in first selection and less than $3$ in second selection.

= $\frac{2}{6}\times \frac{1}{5}+\frac{1}{6}\times \frac{2}{5}=\frac{4}{30}$

$P(X=4)$= larger of two numbers is $4$.

= $\frac{3}{6}\times \frac{1}{5}+\frac{1}{6}\times \frac{3}{5}=\frac{6}{30}$

$P(X=5)$ = larger of two numbers is $5$.

= $\frac{4}{6}\times \frac{1}{5}+\frac{1}{6}\times \frac{4}{5}=\frac{8}{30}$

$P(X=6)$ = larger of two numbers is $6$.

= $\frac{5}{6}\times \frac{1}{5}+\frac{1}{6}\times \frac{5}{5}=\frac{10}{30}$

$\begin{array}{ccccccc}X& 2,& 3& 4& 5& 6& \text{Total}\\ P\left(X\right)& 2/30& 4/30& 6/30& 8/30& 10/30& \\ P\left(X\right).X& 4/30& 12/30& 24/30& 40/30& 60/30& 140/30\\ P\left(X\right).X^2& 8/30& 36/30& 96/30& 200/30& 360/30& 700/30\end{array}$

Mean = $\sum P\left(X\right).X=\frac{140}{30}=\frac{14}{3}$

Variance = $\sum P\left(X\right).X^2-(\sum P(X).X{)}^2$

= $\sum P\left(X\right).X^2-(Mean{)}^2$

= $\frac{70}{3}-{\left(\frac{14}{3}\right)}^2$

= $\frac{70}{3}-\frac{196}{9}=\frac{210-196}{9}=\frac{14}{9}$