Question
Two cards are selected at random from a box which contains five cards numbered 1, 1, 2, 2, and 3. Let X denote the sum and Y the maximum of the two numbers drawn. Find the probability distribution, mean and variance of X and Y.

Answer

Box contains five cards $1, 1, 2, 2, 3$. Here,
X denotes the sum of the two number on cards drawn. Y denotes the maximum of the two number.
So, $X = 2, 3, 4, 5\ Y = 1, 2, 3\ P(X = 2) = P(1)P(1)$
$=\frac{2}{5}\times\frac{1}{4}$
$=0.1$ $ P(X = 3) = P(1)P(2) + P(2)P(1)$
$=\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}$
$=0.4$ $P(X = 4) = P(2)P(2) + P(1)P(3) + P(3)P(1)$
$=\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.3$ $P(X = 5) = P(2)P(3) + P(3)P(2)$
$=\frac{2}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.2$ Probability distribution for X
$x:$ $2$ $3$ $4$ $5$
$P(x):$ $0.1$ $0.4$ $0.3$ $0.2$
Now,
$x_i$ $p_i$ $p_i$ $x_ip_i$ ${x_i}^2p_i$
$2$ $0.1$ $0.1$ $0.4$
$3$ $0.4$ $1.2$ $3.6$
$4$ $0.3$ $1.2$ $4.8$
$5$ $0.2$ $1.0$ $5.0$
    $\sum\text{xp}=3.6$ $\sum\text{x}^2\text{p}=13.8$
Mean $=\sum\text{xp}$
Mean $= 3.6$
Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$
$=13.8-(3.6)^2$ $=13.8-12.96$
Variance $= 0.84\ P(Y = 1) = P(1)P(1)$
$=\frac{2}{5}\times\frac{1}{4}$
$=\frac{2}{20}$
$=0.1$ $P(Y = 2) = P(1)P(2) + P(2)P(1) + P(2)P(2)$
$=\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{2}{4}+\frac{2}{5}\times\frac{1}{4}$
$=0.5$ $P(Y = 3) = P(1)P(3) + P(2)P(3) + P(3)P(1) + P(3)P(2)$
$=\frac{2}{5}\times\frac{1}{4}+\frac{2}{5}\times\frac{1}{4}+\frac{1}{5}\times\frac{2}{4}+\frac{1}{5}\times\frac{2}{4}$
$=0.4$ Probability distribution for Y is
$x:$
$1$
$2$
$3$
$p(x):$
$0.1$
$0.5$
$0.4$
 
$y_i$ $p_i$ $y_ip_i$ ${y_i}^2p_i$
$1$ $0.1$ $0.1$ $0.1$
$2$ $0.5$ $1.0$ $2.0$
$3$ $0.4$ $1.2$ $3.6$
    $\sum\text{xp}=2.3$ $\sum\text{x}^2\text{p}=5.7$
Mean $=\sum\text{xp}=2.3$ Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$ $=5.1-(2.3)^2$ Variance $= 0.41$

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