Question

Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X. [CBSE 2015]

Answer 1

As, X denote the larger of the two numbers obtained.

So, X can take the values 3, 4, 5, 6 and 7.

Now,

$$\begin{gathered} \mathrm{P}\left(X=3\right)=\mathrm{P}\left\{\left(2,3\right)\mathrm{or}\left(3,2\right)\right\}=2\times \frac{1}{6}\times \frac{1}{5}=\frac{1}{15}, \\ \mathrm{P}\left(X=4\right)=\mathrm{P}\left\{\left(2,4\right)\mathrm{or}\left(4,2\right)\mathrm{or}\left(3,4\right)\mathrm{or}\left(4,3\right)\right\}=4\times \frac{1}{6}\times \frac{1}{5}=\frac{2}{15}, \\ \mathrm{P}\left(X=5\right)=\mathrm{P}\left\{\left(2,5\right)\mathrm{or}\left(5,2\right)\mathrm{or}\left(3,5\right)\mathrm{or}\left(5,3\right)\mathrm{or}\left(4,5\right)\mathrm{or}\left(5,4\right)\right\}=6\times \frac{1}{6}\times \frac{1}{5}=\frac{1}{5}, \\ \mathrm{P}\left(X=6\right)=\mathrm{P}\left\{\left(2,6\right)\mathrm{or}\left(6,2\right)\mathrm{or}\left(3,6\right)\mathrm{or}\left(6,3\right)\mathrm{or}\left(4,6\right)\mathrm{or}\left(6,4\right)\mathrm{or}\left(5,6\right)\mathrm{or}\left(6,5\right)\right\}=8\times \frac{1}{6}\times \frac{1}{5}=\frac{4}{15}, \\ \mathrm{P}\left(X=7\right)=\mathrm{P}\left\{\left(2,7\right)\mathrm{or}\left(7,2\right)\mathrm{or}\left(3,7\right)\mathrm{or}\left(7,3\right)\mathrm{or}\left(4,7\right)\mathrm{or}\left(7,4\right)\mathrm{or}\left(5,7\right)\mathrm{or}\left(7,5\right)\mathrm{or}\left(6,7\right)\mathrm{or}\left(7,6\right)\right\}=10\times \frac{1}{6}\times \frac{1}{5}=\frac{1}{3} \end{gathered}$$

The probability distribution of X is as follows:

X:	3	4	5	6	7
P(X):	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{4}{15}$	$\frac{1}{3}$

$$\begin{gathered} \mathrm{So},\mathrm{Mean},\mathrm{E}\left(X\right)=3\times \frac{1}{15}+4\times \frac{2}{15}+5\times \frac{1}{5}+6\times \frac{4}{15}+7\times \frac{1}{3} \\ =\frac{3}{15}+\frac{8}{15}+\frac{15}{15}+\frac{24}{15}+\frac{35}{15} \\ =\frac{85}{15} \\ =\frac{17}{3} \\ \mathrm{Also},\mathrm{E}\left(X^2\right)=3^2\times \frac{1}{15}+4^2\times \frac{2}{15}+5^2\times \frac{1}{5}+6^2\times \frac{4}{15}+7^2\times \frac{1}{3} \\ =\frac{9}{15}+\frac{32}{15}+\frac{75}{15}+\frac{144}{15}+\frac{245}{15} \\ =\frac{505}{15} \\ =\frac{101}{3} \\ \mathrm{Var}\left(X\right)=\mathrm{E}\left(X^2\right)-{\left[\mathrm{E}\left(X\right)\right]}^2 \\ =\frac{101}{3}-{\left(\frac{17}{3}\right)}^2 \\ =\frac{101}{3}-\frac{289}{9} \\ =\frac{303-289}{9} \\ =\frac{14}{9} \end{gathered}$$