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 E(X).

Answer 1

The two positive integers can be selected from the first six positive integers without replacement in $6\times 5=30$ways.

The random variable X may have values2,3,4,5 or 6

$(\because$ 1 cannot be greater than the other selected number)

P(X=2)=P[2 and a number less than 2 are selected or (1,2),(2,1)]

$=\frac{2}{30}=\frac{1}{15}$

P(X=3)=P[3 and a number less than 3 are selected or (1,3),(2,3),(3,1)and (3,2)] =$\frac{4}{30}=\frac{2}{15}$

P(X=4)=P[4 and a number less than 4 are selected or (1,4),(2,4),(3,4),(4,1),(4,2),(4,3)]

=$\frac{6}{30}=\frac{3}{15}$

P(X=5)=P[5 and a number less than 5 are selected or (1,5),(2,5),(3,5),(4,5),(5,1)(5,2),(5,3),(5,4)]

=$\frac{8}{30}=\frac{4}{15}$

P(X=6)=P[6 and a number less than 5 are selected or (1,6),(2,6),(3,6),(4,6),(5,6),(6,1)(6,2),(6,3),(6,4),(6,5)]

=$\frac{10}{30}=\frac{1}{3}$

Therefore, the required probability distribution is as follows

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

$\because$ Expectation of X = E(X)=$\sum XP\left(X\right)$

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

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