On throwing a dice a number greater than four on the die is $5,6$.
$\therefore$ Probability of success $(p)=\frac{2}{6}=\frac{1}{3}$
$\therefore q=1-p=1-\frac{1}{3}=\frac{2}{3}$
Now, variance of probability distribution
$=n p q (\because n=2) $
$=2 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{9} $