Two cards are selected at random from a box which contains five c… - Vidyadip

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 $Y.$

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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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