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 $E(X)$ .

.: Let sample space be $S = \{ (1,2),(1,3),(2,3),(1,4),(2,4),(3,4),(1,5),(2,5),(3,5)$

$$\left( {4,5} \right),\left( {1,6} \right),\left( {2,6} \right),\left( {3,6} \right),\left( {4,6} \right),\left( {5,6} \right)$$

Let $X$ denote the random variable which represents the larger of the two number from first six positive integers.

$$\therefore $$ $X$ can assume values 2, 3, 4, 5, 6

[ 1 can't be greater than any other selected number]

$$\therefore $$ $P(X = 2) = P(1,2) = P(2{\rm{ and anumber less than }}2) = \cfrac{1}{{15}}$

$P(X = 3) = P((1,3),(2,3)) = \cfrac{2}{{15}}$

$P(X = 4) = P((1,4),(2,4),(3,4)) = \cfrac{3}{{15}}$

$P(X = 5) = P((1,5),(2,5),(3,5),(4,5)) = \cfrac{4}{{15}}$

and ${\rm{P}}(X = 6) = P((1,6)(2,6),(3,6),(4,6),(5,6)) = \cfrac{5}{{15}}$

Hence, the probability distribution is

$E(X) = \sum X P(X) = 2 \times \cfrac{1}{{15}} + 3 \times \cfrac{2}{{15}} + 4 \times \cfrac{3}{{15}} + 5 \times \cfrac{4}{{15}} + 6 \times \cfrac{5}{{15}}$

$= \cfrac{1}{{15}}(2 + 6 + 12 + 20 + 30) = \cfrac{{70}}{{15}} = \cfrac{{14}}{3}$