If a random variable X has the probability distribution given by $\mathrm{P}(\mathrm{X}=0)=3{\mathrm{C}}^3,$ $\mathrm{P}(\mathrm{X}=2)=5\mathrm{C}-10{\mathrm{C}}^2$ and $\mathrm{P}(\mathrm{X}=4)=4\mathrm{C}-1.$ then the variance of that distribution is 
 

Given that 

Given $\mathrm{P}(\mathrm{X}=0)=3{\mathrm{C}}^3, \mathrm{P}(\mathrm{X}=2)=5\mathrm{C}-10{\mathrm{C}}^2, \mathrm{P}(\mathrm{X}=4)=4\mathrm{C}-1$

Since  $\sum \mathrm{P}(\mathrm{X}=\mathrm{x})=1$

$$\begin{gathered} \Rightarrow 3{\mathrm{C}}^3+5\mathrm{C}-10{\mathrm{C}}^2+4\mathrm{C}-1=1 \\ \Rightarrow 3{\mathrm{C}}^3-10{\mathrm{C}}^2+9\mathrm{C}-2=0 \\ \Rightarrow (\mathrm{C}-1) (3{\mathrm{C}}^2-7\mathrm{C}+2)=0 \\ \Rightarrow (\mathrm{C}-1) (3\mathrm{C}-1) (\mathrm{C}-2)=0 \\ \Rightarrow \mathrm{C}=1, \frac{1}{3}, 2 \\ \Rightarrow \mathrm{C}=\frac{1}{3} \left(\because 0\le \mathrm{P}\le 1\right) \end{gathered}$$

Mean $\left(\overline{\mathrm{x}}\right)=\sum {\mathrm{x}}_{\mathrm{i}} \mathrm{P}(\mathrm{X}= {\mathrm{x}}_{\mathrm{i}})$

                $=0+\frac{10}{9}+\frac{4}{3}=\frac{22}{9}$

Now Variance $\left({\mathrm{a}}^2\right)=\sum {({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}})}^2.\mathrm{P}(\mathrm{X}={\mathrm{x}}_{\mathrm{i}})$

                                 $$\begin{gathered} ={\left(0-\frac{22}{9}\right)}^2 \left(\frac{1}{9}\right)+{\left(2-\frac{22}{9}\right)}^2 \left(\frac{5}{9}\right)+{\left(4-\frac{22}{9}\right)}^2 \left(\frac{1}{3}\right) \\ =\frac{1152}{729}=\frac{128}{81} \end{gathered}$$