A die is rolled twice. Getting a number greater than 4 is called a ‘Success’. Let the random variable X is equal to the number of successes then mean of X is 
 

Let X be a radom variable of number of success.

 So, X can take values 0, 1, 2

 let 'S' denote the success of  getting 5 or 6

$$\begin{gathered} \therefore \mathrm{P}\left(\mathrm{S}\right)=\mathrm{P}(5 \mathrm{or} 6)=\frac{2}{6}=\frac{1}{3} \\ \mathrm{P}\left(\overline{\mathrm{S}}\right)=1-\frac{1}{3}=\frac{2}{3} \end{gathered}$$

Now $\mathrm{P}(\mathrm{x}=0)=\mathrm{P}(\overline{\mathrm{S}}·\overline{\mathrm{S}})=\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}$

         $$\begin{gathered} \mathrm{P}(\mathrm{x}=1)=\mathrm{P}(\mathrm{S}·\overline{\mathrm{S}})+\mathrm{P}(\overline{\mathrm{S}}·\mathrm{S})=\frac{4}{9} \\ \mathrm{P}(\mathrm{x}=2)=\mathrm{P}(\mathrm{S}·\mathrm{S})=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9} \end{gathered}$$

Mean $=0 \left(\frac{4}{9}\right)+1 \left(\frac{4}{9}\right)+2 \left(\frac{1}{9}\right)$

           $$\begin{gathered} =\frac{6}{9} \\ =\frac{2}{3} \end{gathered}$$