# 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.

$(a)\;\large\frac{17}{3} \qquad(b)\;\large\frac{14}{3}\qquad(c)\;\large\frac{15}{2}\qquad(d)\;\large\frac{17}{5}$

The numbers are 1,2,3,4,5,6
Let X be the bigger number among the two number.
i.e., X = 2,3,4,5,6
$P(x = 2) = P(1 , 2)$
$= \large\frac{1}{6C_2}$$= \large\frac{1}{15} P(x = 3) = P(1 , 3) (2,3) = \large\frac{2}{6C_2}$$= \large\frac{2}{15}$
$P(x = 4) = P(1 , 4) (2,4) (3,4)$
$= \large\frac{3}{6C_2}$$= \large\frac{3}{15}$$= \large\frac{1}{5}$
$P(x = 5) = P(1 , 5) (2,5) (3,5) (4,5)$
$= \large\frac{4}{6C_2}$$=\large\frac{4}{15} P(x = 6) = P(1 , 6) (2,6) (3,6) (4,6) (5,6) = \large\frac{5}{6C_2}$$= \large\frac{5}{15}$$= \large\frac{1}{3} Hence the probability distribution is : X \quad\quad \quad \quad2\quad\quad 3 \quad \: \: 4\quad \: \: 5 \: \: \quad 6 p(x=x) \: \: \large\frac{1}{15} \quad \large\frac{2}{15} \quad \large\frac{1}{5} \quad \large\frac{4}{15} \quad \large\frac{1}{3} Mean = 2 \bigg(\large\frac{1}{15} \bigg)+3 \bigg(\large\frac{2}{15} \bigg) + 4 \bigg(\large\frac{1}{5} \bigg) + 5 \bigg (\large\frac{4}{15} \bigg)+ 6 \bigg (\large\frac{1}{3} \bigg) \large\frac{2}{15}$$+ \large\frac{6}{15}$$+ \large\frac{4}{5}$$ + \large\frac{20}{15}$$+ \large\frac{6}{3} \large\frac{46}{15}$$+ \large\frac{4}{5}$$+ 2 \large\frac{70}{15}$$ \large\frac{14}{3}$
Hence the mean is $\large\frac{14}{3}$