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,
$x_i$ $p_i$ $x_ip_i$ $x_i^2p_i$
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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