An urn contains 5 red and 2 blcak balls. Two balls are randomly d… - Vidyadip

Question

An urn contains 5 red and 2 blcak balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, then find the mean and variance of X.

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Answer

X can assume values 0 to 2. Yes X is a random variable. P(X = 0) = (Probability of getting no black ball) $=\frac{\text{}^{2}\text{C}_0\times\text{}^{5}\text{C}_2}{\text{}^{7}\text{C}_2}=\frac{1\times\frac{5\times4}{2\times1}}{\frac{7\times6}{2\times1}}=\frac{20}{42}$ P(X = 1) = (Probability of getting one black ball) $=\frac{\text{}^{2}\text{C}_1\times\text{}^{5}\text{C}_1}{\text{}^{7}\text{C}_2}=\frac{1\times5}{\frac{7\times6}{2\times1}}=\frac{20}{42}$ P(X = 2) = (Probability of getting two black balls) $=\frac{\text{}^{2}\text{C}_2\times\text{}^{5}\text{C}_0}{\text{}^{7}\text{C}_2}=\frac{1\times1}{\frac{7\times6}{2\times1}}=\frac{2}{42}$ Thus, Probability distribution of random variable X is

$\text{X}$	$0$	$1$	$2$
$\text{P}(\text{X}) $	$\frac{20}{42}$	$\frac{20}{42}$	$\frac{2}{42}$

$\text{x}_\text{i}$	$\text{p}_\text{i}$	$\text{p}_\text{i}\text{x}_\text{i}$	$\text{p}_\text{i}\text{X}_\text{i}^2$
$0$	$\frac{20}{42}$	$0$	$0$
$1$	$\frac{20}{42}$	$\frac{20}{42}$	$\frac{20}{42}$
$2$	$\frac{2}{42}$	$\frac{4}{42}$	$\frac{8}{42}$
		$\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$	$\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{2}{3}$

Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=\frac{4}{7}$
Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2$
$=\frac{2}{3}-\Big(\frac{4}{7}\Big)^2=\frac{50}{147}$

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