There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let $X$ denotes the sum of the numbers on two cards drawn. Find the mean and variance of $X$.
Here, $$S=\{(1,2),(2,1),(1,3),(3,1),(2,3),(3,2),(1,4),(4,1),(1,5),(5,1),(2,4),(4,2),(2,5),(5,2),(3,4),(4,3),(3,5),(5,3),(5,4),(4,5)\}$$
$$\Rightarrow n(S)=20$$
Let random variable be $X$ which denotes the sum of the numbers on two cards drawn.
$\therefore\qquad X=3,4,5,6,7,8,9$
$$\begin{aligned} & \text { At } X=3, P(X)=\frac{2}{20}=\frac{1}{10} \\ & \text { At } X=4, P(X)=\frac{2}{20}=\frac{1}{10} \\ & \text { At } X=5, P(X)=\frac{4}{20}=\frac{1}{5} \\ & \text { At } X=6, P(X)=\frac{4}{20}=\frac{1}{5} \\ & \text { At } X=7, P(X)=\frac{4}{20}=\frac{1}{5} \\ & \text { At } X=8, P(X)=\frac{2}{20}=\frac{1}{10} \\ & \text { At } X=9, P(X)=\frac{2}{20}=\frac{1}{10} \end{aligned}$$
$\therefore \quad$ Mean, $E(X)=\Sigma X P(X)=\frac{3}{10}+\frac{4}{10}+\frac{5}{5}+\frac{6}{5}+\frac{7}{5}+\frac{8}{10}+\frac{9}{10}$
$=\frac{3+4+10+12+14+8+9}{10}=6$
$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} \Sigma X^2 P(X) & =\frac{9}{10}+\frac{16}{10}+\frac{25}{5}+\frac{36}{5}+\frac{49}{5}+\frac{64}{10}+\frac{81}{10} \\ & =\frac{9+16+50+72+98+64+81}{10}=39 \end{aligned} \end{aligned}$$
$$\begin{aligned} \therefore \quad \operatorname{Var}(X) & =\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =39-(6)^2=39-36=3 \end{aligned}$$
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