When a die is rolled twice, if the event of getting an even number is denoted by a success and the number of successes as a random
variable, then distribution and mean of the variate are

Let X be the random Variable for a success for getting an even number on a toss.

$\therefore \mathrm{x}=0, 1, 2$ and $\mathrm{n}=2$,

$\mathrm{p}=\frac{3}{2}=\frac{1}{2} \Rightarrow \mathrm{q}=1-\mathrm{p}=\frac{1}{2}$

At $\mathrm{X}=0, \mathrm{P}(\mathrm{x}=0)={}^2{\mathrm{c}}_0 {\left(\frac{1}{2}\right)}^0 {\left(\frac{1}{2}\right)}^{2-0}=\frac{1}{4}$

At $\mathrm{X}=1, \mathrm{P}(\mathrm{X}=1)={}^2{\mathrm{c}}_1 {\left(\frac{1}{2}\right)}^1 {\left(\frac{1}{2}\right)}^{2-1}$

                                  $$\begin{gathered} =2\times \frac{1}{2}\times \frac{1}{2} \\ =\frac{1}{2} \end{gathered}$$

At $\mathrm{X}=2, \mathrm{P}(\mathrm{X}=2)={}^2{\mathrm{c}}_2 {\left(\frac{1}{2}\right)}^2 {\left(\frac{1}{2}\right)}^{2-2}$

                                   $$\begin{gathered} =1\times \frac{1}{4}\times 1 \\ =\frac{1}{4} \end{gathered}$$

$\therefore$ Probability distribution of X is

$\begin{array}{rcl}\therefore \mathrm{Mean}& =& \sum \mathrm{X}.\mathrm{P}(\mathrm{X}=\mathrm{x})\\ & =& 0\left(\frac{1}{4}\right)+1\left(\frac{1}{2}\right)+2\left(\frac{1}{4}\right)\\ & =& \frac{1}{2}+\frac{1}{2}\\ & =& 1\end{array}$