Question

There are $4$ cards numbered $1,3,5\text{and}7,$ 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

Clearly, $X$ can take values as $4,6,8,10\text{and}12$
$P(X=4)=P\left(\text{getting 1 in first draw and 3 in}{2}^{nd}\text{draw}\right)or\left(\text{getting 3 in first draw and 1 in}{2}^{nd}\text{draw}\right)$ $=\frac{1}{4}\times \frac{1}{3}+\frac{1}{4}\times \frac{1}{3}=\frac{2}{12}=\frac{1}{6}P(X=6)=P\left(\text{getting 1 in first draw and 5 in}{2}^{nd}\text{draw}\right)or\left(\text{getting 5 in first draw and 1 in}{2}^{nd}\text{draw}\right)=\frac{1}{4}\times \frac{1}{3}+\frac{1}{4}\times \frac{1}{3}=\frac{2}{12}=\frac{1}{6}\text{Similarly},P(X=8)=\frac{1}{4}\times \frac{1}{3}+\frac{1}{4}\times \frac{1}{3}+\frac{1}{4}\times \frac{1}{3}+\frac{1}{4}\times \frac{1}{3}=\frac{4}{12}=\frac{1}{3}P(X=10)=\frac{1}{4}\times \frac{1}{3}+\frac{1}{4}\times \frac{1}{3}=\frac{2}{12}=\frac{1}{6}P(X=12)=\frac{1}{4}\times \frac{1}{3}+\frac{1}{4}\times \frac{1}{3}=\frac{2}{12}=\frac{1}{6}$ Thus, the probability distribution of $X$ is as: $\begin{array}{cccccc}X:& 4& 6& 8& 10& 12\\ \text{P(X)}:& \frac{1}{6}& \frac{1}{6}& \frac{1}{3}& \frac{1}{6}& \frac{1}{6}\\ \text{XP(X)}:& \frac{4}{6}& \frac{6}{6}& \frac{8}{3}& \frac{10}{6}& \frac{12}{6}\\ {\text{X}}^2\text{P}\left(X\right):& \frac{16}{6}& \frac{36}{6}& \frac{64}{3}& \frac{100}{6}& \frac{144}{6}\end{array}\therefore Mean=\overline{X}=\sum XP\left(X\right)=\frac{4}{6}+\frac{6}{6}+\frac{8}{3}+\frac{10}{6}+\frac{12}{6}=\frac{4+6+16+10+12}{6}=\frac{48}{6}=8Var\left(X\right)=\sum X^{2}P\left(X\right)-{(\sum XP(X\left)\right)}^2=\frac{16}{6}+\frac{36}{6}+\frac{64}{3}+\frac{100}{6}+\frac{144}{6}-{\left(8\right)}^2=\frac{16+36+128+100+144}{6}-64=\frac{424}{6}-64=\frac{424-384}{6}=\frac{40}{6}=\frac{20}{3}=6.67$