When a die is tossed up twice, we get 6 $\times$ 6 = 36 outcomes. Given that success is defined as any number greater than 4.

The sample space of the roll of the die is given by: S = $\begin{Bmatrix} 1,1 &1,2 & 1,3 & 1,4& 1,5 &1,6 \\ 2,1 &... & & & & \\ 3,1 &... & & & & \\ 4,1 &... & & & & \\ 5,1&... & & & & \\ 6,1 & ... & & & & 6,6 \end{Bmatrix}$

(i) Let X be the random variable that defines the number of tosses greater than 4.

P (X = 0) = P (number $\leq$ 4 on both tosses) = $\begin{Bmatrix} 1,1 &1,2 & 1,3 & 1,4\\ 2,1 &... & & \\ 3,1 &... & & \\ 4,1 & ... & & 4,4 \end{Bmatrix}$ = $\large\frac{16}{36} = \frac{4}{9}$

P (X = 1) = P (number $\leq$ 4 on only one of the tosses and $\geq$ on the second toss) = $\begin{Bmatrix} 1,5 &1,6 \\ 2,5 &2,6 \\ 3,5 &3,6 \\ 4,5 &4,6\\ 5,1&5,2&5,3&5,4\\ 6,1&6,2&6,3&6,4 \end{Bmatrix}$ = $\large\frac{16}{36} = \frac{4}{9}$

P (X = 2) = P (number greater than 4 on both tosses) = $\begin{Bmatrix} 5,5&5,6\\ 6,5&6,6\\ \end{Bmatrix}$ = $\large\frac{4}{36} = \frac{1}{9}$

Therefore the probability distribution of X can be written as follows:

$\begin{matrix} \textbf{X} & 0 &1 &2 \\ \textbf{P(X)} &\large \frac{4}{9} &\large \frac{4}{9} &\large \frac{1}{9} \end{matrix}$

(ii) Let X be the random variable that defines the number of tosses where 6 appears on at least one die.

P (X = 0) = P (6 does not appear on either die) = $\begin{Bmatrix} 1,1 &1,2 & 1,3 & 1,4& 1,5 \\ 2,1 &... & & & \\ 3,1 &... & & & \\ 4,1 &... & & & \\ 5,1&... & & & 5,5 \end{Bmatrix}$ = $\large\frac{25}{36}$

P (X = 1) = P (6 appears at least on one die) = $\begin{Bmatrix} 1,6&2,6&3,&4,6&5,6&6,6\\ 6,1&6,2&6,3&6,4&6,5 \end{Bmatrix}$ = $\large\frac{11}{36}$

Therefore the probability distribution is represented as follows: $\begin{matrix} \textbf{X} & 0 &1 \\ \textbf{P(X)} &\large \frac{25}{36} &\large \frac{11}{36} \end{matrix}$