A biased die is such that $P(4) = \frac{1}{{10}}$ and other scores being equally likely. The die is tossed twice. If X is the 'number of fours seen', then find the variance of the random variable X.

Since, $X =$ Number of fours seen
On tossing two die,

$X = 0,1,2$
Also, ${P_{(4)}} = \frac{1}{{10}}$

and ${P_{({\rm{not }}4)}} = \frac{9}{{10}}$
So, $P(X = 0) = {P_{({\rm{not}}4)}} \cdot {P_{({\rm{not}}4)}} = \frac{9}{{10}} \cdot \frac{9}{{10}} = \frac{{81}}{{100}}$

$P(X = 1) = {P_{({\rm{not }}4)}} \cdot {P_{(4)}} + {P_{(4)}} \cdot {P_{({\rm{not}}4)}} = \frac{9}{{10}} \cdot \frac{1}{{10}} + \frac{1}{{10}} \cdot \frac{9}{{10}} = \frac{{18}}{{100}}$

$P(X = 2) = {P_{(4)}} \cdot {P_{(4)}} = \frac{1}{{10}} \cdot \frac{1}{{10}} = \frac{1}{{100}}$

Thus, we get following table

$\therefore$ ${\mathop{\rm Var}\nolimits} (X) = E\left( {{X^2}} \right) - {[E(X)]^2} = \Sigma {X^2}P(X) - {[\Sigma XP(X)]^2}$

$= \left[ {0 + \frac{{18}}{{100}} + \frac{4}{{100}}} \right] - {\left[ {0 + \frac{{18}}{{100}} + \frac{2}{{100}}} \right]^2}$
$= \frac{{22}}{{100}} - {\left( {\frac{{20}}{{100}}} \right)^2} = \frac{{11}}{{50}} - \frac{1}{{25}}$

$= \frac{{11 - 2}}{{50}} = \frac{9}{{50}} = \frac{{18}}{{100}} = 0.18$