Question

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

Answer 1

$\text{X can take values 2, 3, 4, 5, 6}$
$(\because \text{1 cannot be greater than the other selected number)}$

$\therefore P(X=2)=P\text{(2 and a number less than 2 are selected)}$

$=\frac{1}{6_{C_2}}=\frac{1}{\frac{6\times 5}{2}}=\frac{1}{15}$

$P(X=3)=P\text{(3 and a number less than 3 are selected)}=\frac{2}{6_{C_2}}$
$=\frac{2}{15}\left[\right(1,3),(2,3\left)\right]$

$P(X=4)=P\text{(4 and a number less than 4 are selected)}$

$=\frac{3}{6_{C_2}}=\frac{3}{15}[\because (1,4),(2,4),(3,4\left)\right]$

$P(X=5)=P\text{(5 and a number less than 5 are selected}$

$=\frac{4}{6_{C_2}}=\frac{4}{15}[\because (1,5),(2,5),(3,5),(4,5\left)\right]$

$P(X=6)=P\text{(6 and a number less than 6 are selected)}$
$=\frac{5}{6_{C_2}}=\frac{5}{15}[\because (1,6),(2,6),(3,6),(4,6),(5,6\left)\right]$

$\therefore \text{Probability distribution is}$

$\begin{array}{cccccc}\text{X}& 2& 3& 4& 5& 6\\ \text{P(X)}& \frac{1}{15}& \frac{2}{15}& \frac{3}{15}& \frac{4}{15}& \frac{5}{15}\end{array}$


$\text{Mean of}$ $X=E\left(X\right)=\sum XP\left(X\right)$

$=2\times \frac{1}{15}+3\times \frac{2}{15}+4\times \frac{3}{15}+5\times \frac{4}{15}+6\times \frac{5}{15}$

$=\frac{2+6+12+20+30}{15}=\frac{70}{15}=\frac{14}{3}$