If a random variable X has the probability distribution given by

 $P\left(X=0\right)=3C^3,P\left(X=2\right)=5C-10C^2 and P\left(X=4\right) = 4C-1,$

then the variance of that distribution is

Given Probability distribution of X is

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

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

Then Probability distribution is 

Mean  $\left(\mathrm{\mu }\right)=\sum \mathrm{x} \mathrm{P}\left(X\right)$

                 $=0+\frac{10}{9}+\frac{4}{3}=\frac{10+12}{9}=\frac{22}{9}$

Variance $=\sum {\mathrm{x}}^2 \mathrm{P}(\mathrm{X}=\mathrm{x})-{\mathrm{\mu }}^2$

                 $$\begin{gathered} =0+4\left(\frac{5}{9}\right)+16\left(\frac{1}{3}\right)-{\left(\frac{22}{9}\right)}^2 \\ =\frac{68}{9}-\frac{484}{81} \\ =\frac{612-484}{81} \\ =\frac{128}{81} \end{gathered}$$