Question

There are $4$ cards numbered $1,3,5$ and $7$, one number on one card. Two cards are drawn at random without replacement. Let $X$ denote the sum of the numbers on the two drawn cards. Find the mean and variance of $X$.

Answer 1

The $4$ cards are numbered as $1,3,5,7$

Let $S$ be the sample space and we have $4\times 3=12$ elements in sample space

Here $X$ is sum of numbers on two drawn cards , $X$ can be $4,6,8,10,12$

$P(X=4)=\frac{2}{12}=\frac{1}{6}$

$P(X=6)=\frac{2}{12}=\frac{1}{6}$

$P(X=8)=\frac{4}{12}=\frac{1}{3}$

$P(X=10)=\frac{2}{12}=\frac{1}{6}$

$P(X=12)=\frac{2}{12}=\frac{1}{6}$

$\sum P_i{X}_i=4\times \frac{1}{6}+6\times \frac{1}{6}+8\times \frac{1}{3}+10\times \frac{1}{6}+12\times \frac{1}{6}=\frac{4+6+16+10+12}{6}=\frac{48}{6}=8$

Therefore the mean is $8$

$\sum P_i{X_i}^2=4^2\times \frac{1}{6}+6^2\times \frac{1}{6}+8^2\times \frac{1}{3}+{10}^2\times \frac{1}{6}+{12}^2\times \frac{1}{6}=\frac{16+36+128+100+144}{6}=\frac{424}{6}=\frac{212}{3}$

The variance is $\frac{212}{3}-64=\frac{20}{3}$