Find the probability distribution of the number of successes in two tosses of a die,where a success is defined as a number greater than $4$.

(A) When a die is tossed two times,the total number of possible outcomes is $6 \times 6 = 36$.
Let $X$ be the random variable representing the number of successes.
$A$ success is defined as obtaining a number greater than $4$ (i.e.,$5$ or $6$).
The probability of success in a single toss is $p = \frac{2}{6} = \frac{1}{3}$.
The probability of failure in a single toss is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.
$X$ can take values $0, 1, 2$.
$P(X=0) = q \times q = \frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$.
$P(X=1) = (p \times q) + (q \times p) = (\frac{1}{3} \times \frac{2}{3}) + (\frac{2}{3} \times \frac{1}{3}) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}$.
$P(X=2) = p \times p = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$.
The probability distribution is:

$X$	$0$	$1$	$2$
$P(X)$	$\frac{4}{9}$	$\frac{4}{9}$	$\frac{1}{9}$