Q.A fair die is thrown 3 times. If getting "a number greater than 4" is called a success, find the probability distribution of the number of successes and hence find its mean and variance.

"Greater than 4" means rolling a 5 or a 6, so $p=\dfrac{2}{6}=\dfrac{1}{3}$ per throw. With $n=3$ independent throws, the number of successes $X\sim B(3,\,1/3)$ — build the full distribution table from $P(X=r)=\binom{n}{r}p^rq^{n-r}$, then apply mean $=np$, variance $=npq$.

$p=P(\text{5 or 6})=\dfrac{2}{6}=\dfrac{1}{3},\quad q=\dfrac{2}{3},\quad n=3$

$P(X=0)=\binom{3}{0}\left(\dfrac{1}{3}\right)^0\left(\dfrac{2}{3}\right)^3=\dfrac{8}{27}$

$P(X=1)=\binom{3}{1}\left(\dfrac{1}{3}\right)^1\left(\dfrac{2}{3}\right)^2=3\times\dfrac{1}{3}\times\dfrac{4}{9}=\dfrac{4}{9}$

$P(X=2)=\binom{3}{2}\left(\dfrac{1}{3}\right)^2\left(\dfrac{2}{3}\right)^1=3\times\dfrac{1}{9}\times\dfrac{2}{3}=\dfrac{2}{9}$ …