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.

X can assume values 0 to 2. Yes X is a random variable. P(X = 0) = (Probability of getting no black ball) = ^ 2 C_0 ^ 5 C_2 ^ 7 C_2 = 1 5 4 2 1 7 6 2 1 =20 42 P(X = 1) = (Probability of getting one black ball) = ^ 2 C_1 ^ 5 C_1 ^ 7 C_2 =1 5 7 6 2 1 =20 42 P(X = 2) = (Probability of getting two black balls) = ^ 2 C_2 ^ 5 C_0 ^ 7 C_2 =1 1 7 6 2 1 =2 42 Thus, Probability distribution of random variable X is X 0 1 2 P(X) 20 42 20 42 2 42 x_i p_i p_ix_i p_iX_i^2 0 20 42 0 0 1 20 42 20 42 20 42 2 2 42 4 42 8 42 _ix_i=4 7 _ix_i^2=2 3 Mean = _ix_i=4 7 Variance = _ix_i^2- ( _ix_i )^2 =2 3 - (4 7 )^2=50 147

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

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

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

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