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

Question

An urn contains 5 red and 2 black 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, find the mean and variance of X.

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Answer

Possible values of x are 0, 1, 2 and x is a random variable:

$\text{x:}$	$\text{P(x)}$	$\text{x P(x)}$	$\text{x}^{2} \text{P(x)}$
0	$\frac{^2{\text{C}_{0}}\times^5{\text{C}_{2}}}{^7{\text{C}_{2}}} = \frac{20}{42}$	0	0	$\text{For P (x)}$
1	$\frac{^2{\text{C}_{1}}\times^5{\text{C}_{1}}}{^7{\text{C}_{2}}} = \frac{20}{42}$	$\frac{20}{40}$	$\frac{20}{42}$	$\text{For x P (x)}$
2	$\frac{^2{\text{C}_{2}}\times^5{\text{C}_{0}}}{^7{\text{C}_{2}}} = \frac{2}{42}$	$\frac{4}{42}$	$\frac{8}{42}$	$\text{For x}^{2}\text{ P (x)}$

$\sum \text{x P(x)} = \frac{24}{42}; \sum \text{x}^{2} \text{ P(x)} = \frac{28}{42}$ $\text{Mean} = \sum \text{x P(x)} = \frac{4}{7}; \text{variance} = \sum \text{x}^{2} \text{P(x)} - \bigg[\sum \text{x P (x)}\bigg]^{2}$ $\text{Variance} = \frac{50}{147} =\frac{2}{3}- \frac{16}{49} = \frac{50}{147} $

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