Find the probabilty distribution of the number of successes in two tosses of a dice, where a success is defined as
number greater than 4
six appears on at least one dice

.: Let $S$ be the sample space: $S = \{ 1,2,3,4,5,6\}$

(i) $p =$ probability of getting a number greater than 4 i.e., $\{ 5,6\} = \cfrac{2}{6} = \cfrac{1}{3}$

$q =$ probability of not getting a number greater than 4 i.e.

, $\{ 1,2,3,4\} = \cfrac{4}{6} = \cfrac{2}{3}$

Let $X$ denote the random variables which represents the number comes on the dice will be greater than 4.

$X$ can assume the values 0, 1, 2

$P(X = 0) = q \times q = \cfrac{2}{3} \times \cfrac{2}{3} = \cfrac{4}{9}$

Hence, the probability distribution is :

$P(X = 1) = p \times q + q \times p = \cfrac{1}{3} \times \cfrac{2}{3} + \cfrac{2}{3} \times \cfrac{1}{3} = \cfrac{4}{9}$

$P(X = 2) = p \times p = \cfrac{1}{3} \times \cfrac{1}{3} = \cfrac{1}{9}$

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

(ii) $X$ denote the random variable which represents at least one dice shows six.

$\therefore$ $X$ can assume the values 0, 1, 2
$P(X = 0) = P({\rm{ six does not appears at any dice }}) = \cfrac{5}{6} \times \cfrac{5}{6} = \cfrac{{25}}{{36}}$

$P(X = 1) = P({\rm{ six appears on one dice }}) = \cfrac{5}{6} \times \cfrac{1}{6} + \cfrac{1}{6} \times \cfrac{5}{6} = \cfrac{{10}}{{36}}$

$P(X = 2) = P({\rm{ six appears on both dice }}) = \cfrac{1}{6} \times \cfrac{1}{6} = \cfrac{1}{{36}}$

Hence, the probability distribution :

$X$ 0 1 2
$P(X)$ $\cfrac{{25}}{{36}}$ $\cfrac{{10}}{{36}}$ $\cfrac{1}{{36}}$