A bag contains 4 white and 6 black balls. Three balls are drawn at...

Given,

$X =$ Number of white ball drawn,

And bag contains $4$ white and $6$ black balls, total $3$ balls are to be drawn,

So, $P\left(X = 0\right) = \dfrac{C3 6}{C3 10} = \dfrac{1}{6}$

$P\left(X = 1\right) = \dfrac{C2 6 \times C1 4}{C3 10} = \dfrac{1}{2}$

$P\left(X = 2\right) = \dfrac{C1 6 \times C2 4}{C3 10} = \dfrac{3}{10}$

and $P\left(X = 3\right) = \dfrac{C0 6 \times C3 4}{C3 10} = \dfrac{1}{30}$

$X$	$0$	$1$	$2$	$3$
$P\left(X\right)$	$\dfrac{1}{6}$	$\dfrac{1}{2}$	$\dfrac{3}{10}$	$\dfrac{1}{30}$

So, variance $= {\sigma}^{2} = ∑ {P}_{i}{X}_{i}^{2} - {\left(∑ {P}_{i}{X}_{i}\right)}^{2}$

$⇒ {\sigma}^{2} = \dfrac{1}{2} + \dfrac{12}{10} + \dfrac{3}{10} - {\left(\dfrac{1}{2} + \dfrac{6}{10} + \dfrac{1}{10}\right)}^{2}$

$⇒ {\sigma}^{2} = \dfrac{56}{100}$

$⇒ 100{\sigma}^{2} = 56$