A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If $σ^{2}$ is the variance of X, then 100 $σ^{2}$ is equal to

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answer is 56.

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Detailed Solution

X	0	1	2	3
P(X)	$\frac{^{6} C_{3}}{^{10} C_{3}} = \frac{1}{6}$	$\frac{^{4} C_{1} \times^{6} C_{2}}{^{10} C_{3}} = \frac{1}{2}$	$\frac{^{4} C_{2} \times^{6} C_{1}}{^{10} C_{3}} = \frac{3}{10}$	$\frac{^{4} C_{3}}{^{10} C_{3}} = \frac{1}{30}$

$σ^{2} = {\underset{}{\sum X^{2} P (X) - (\underset{}{\sum X P (X)})}}^{2} = \frac{56}{100} \Rightarrow σ^{2} = (0 + 1 \times \frac{1}{2} + 4 \times \frac{3}{10} + 9 \times \frac{1}{30}) - {(0 + 1 \times \frac{1}{2} + 2 \times \frac{3}{10} + 3 \times \frac{1}{30})}^{2} \Rightarrow σ^{2} = (\frac{1}{2} + \frac{6}{5} + \frac{3}{10}) - {(\frac{1}{2} + \frac{3}{5} + \frac{1}{10})}^{2} = (\frac{20}{10}) - {(\frac{12}{10})}^{2} = 2 - (\frac{144}{100}) = \frac{56}{100} 100 σ^{2} = 56$