A die is tossed twice. If a 'success' is getting an even number on a toss, then find the variance of the number of successes.

Let $X$ be the random variable for a 'success' for getting an even number on a toss.

$\therefore$ $X = 0,1,2,n = 2,p = \frac{3}{6} = \frac{1}{2}$

and $q = \frac{1}{2}$

At $X = 0,\quad P(X = 0){ = ^2}{C_0}{\left( {\frac{1}{2}} \right)^0}{\left( {\frac{1}{2}} \right)^{2 - 0}} = \frac{1}{4}$

At $X = 1,\quad P(X = 1){ = ^2}{C_1}{\left( {\frac{1}{2}} \right)^1}{\left( {\frac{1}{2}} \right)^{2 - 1}} = 2 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2}$

At $X = 2,\quad P(X = 2){ = ^2}{C_2}{\left( {\frac{1}{2}} \right)^2}{\left( {\frac{1}{2}} \right)^{2 - 2}} = \frac{1}{4}$

Thus,

1
$\therefore$ $\quad \Sigma XP(X) = 0 + \frac{1}{2} + \frac{1}{2} = 1$ …….(i)

and $\Sigma {X^2}P(X) = 0 + \frac{1}{2} + 1 = \frac{3}{2}$ ………(ii)

$= \Sigma {X^2}P(X) - {[\Sigma XP(X)]^2} = \frac{3}{2} - {(1)^2} = \frac{1}{2}\quad$ [using Eqs. (i) and (ii)]