A die is thrown twice. If getting a number greater than four on the die is considered a success, then the variance of the probability distribution of the number of successes is
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} $