Question

Two the 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

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

$P(x=2)=P(1,2)$

$⟹=\frac{1}{{}^6{C}_2}=\frac{1}{15}$

$P(x=3)=P(1,3)(2,3)$

$⟹=\frac{2}{{}^6{C}_2}=\frac{2}{15}$

$P(x=4)=P(1,4)(2,4)(3,4)$

$⟹=\frac{3}{{}^6{C}_2}=\frac{3}{15}=\frac{1}{5}$

$P(x=5)=P(1,5)(2,5)(3,5)(4,5)$

$⟹=\frac{4}{{}^6{C}_2}=\frac{4}{15}$

$P(x=6)=P(1,6)(2,6)(3,6)(4,6)(5,6)$

$⟹=\frac{5}{{}^6{C}_2}=\frac{5}{15}=\frac{1}{3}$

Hence the probability distribution is :

X	2	3	4	5	6
p(X=x)	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{4}{15}$	$\frac{1}{3}$

Mean = $2\left(\frac{1}{15}\right)+3\left(\frac{2}{15}\right)+4\left(\frac{1}{5}\right)+5\left(\frac{4}{15}\right)+6\left(\frac{1}{3}\right)$

$⟹=\frac{2}{15}+\frac{6}{5}+\frac{10}{3}$

$⟹=\frac{2}{15}+\frac{68}{15}⟹\frac{70}{15}$

Hence, Mean = $\frac{14}{3}$

Using Variance formula,

$Var\left(X\right)=(X-\mu {)}^2\times P(X=x)⟹Var(X)={(2-\frac{14}{3})}^2\times \frac{1}{15}+{(3-\frac{14}{3})}^2\times \frac{2}{15}+{(4-\frac{14}{3})}^2\times \frac{1}{5}+{(5-\frac{14}{3})}^2\times \frac{4}{15}+{(6-\frac{14}{3})}^2\times \frac{1}{3}⟹Var(X)={(-\frac{8}{3})}^2\times \frac{1}{15}+{(-\frac{5}{3})}^2\times \frac{2}{15}+{(-\frac{2}{3})}^2\times \frac{1}{5}+{\left(\frac{1}{3}\right)}^2\times \frac{4}{15}+{\left(\frac{4}{3}\right)}^2\times \frac{1}{3}⟹Var(X)=\frac{64}{9\times 15}+\frac{50}{9\times 15}+\frac{4}{9\times 5}+\frac{4}{9\times 15}+\frac{16}{9\times 3}⟹Var(X)=\frac{64+50+12+4+80}{9\times 15}=\frac{210}{135}=\frac{14}{9}$

Hence, the variance is $\frac{14}{9}$