A die is thrown twice. If getting a number greater than four on the die is considered a success, then the variance of the probability distribution of the number of successes is

When a die is tossed twice total possible out comes =36

Getting a number greater than '4' is considered as success

$$\begin{gathered} \mathrm{P}(\mathrm{X}=0)=\frac{16}{36}=\frac{4}{9}(\because zero numbers greater than 4=16) \\ \mathrm{P}(\mathrm{X}=1)=\frac{16}{36}=\frac{4}{9}(\because one number greater than 4=16) \\ \mathrm{P}(\mathrm{X}=2)=\frac{4}{36}=\frac{1}{9}(\because two numbers greater than 4=4) \end{gathered}$$

Mean $\left(\mathrm{\mu }\right)=\sum _{ }\mathrm{xP}\left(\mathrm{x}\right)$

                 $$\begin{gathered} =0+\frac{4}{9}+2\left(\frac{1}{9}\right) \\ =\frac{6}{9}=\frac{2}{3} \end{gathered}$$

Now Variance $\left({\sigma }^2\right)=\sum {{\mathrm{x}}_{\mathrm{i}}}^2 \mathrm{P}\left({\mathrm{x}}_{\mathrm{i}}\right)-{\left(\mathrm{\mu }\right)}^2$

                                 $$\begin{gathered} =0+{\left(1\right)}^2 \left(\frac{4}{9}\right)+{\left(2\right)}^2 \left(\frac{1}{9}\right)-{\left(\frac{2}{3}\right)}^2 \\ =\frac{4}{9}+\frac{4}{9}-\frac{4}{9} \\ =\frac{4}{9} \end{gathered}$$