$\text{ For a binomial variate }\mathrm{X}\text{ with }\mathrm{n}=6\text{, if }\mathrm{P}(\mathrm{x}=2)=9\mathrm{P}(\mathrm{x}=4), \text{then its variance is }$

$$\begin{gathered} \text{ Given, }\mathrm{n}=6\text{ and } \\ \begin{array}{l}\mathrm{P}(\mathrm{x}=2)=9\mathrm{P}(\mathrm{X}=4)\\ {\Rightarrow }^6{\mathrm{C}}_2{\mathrm{p}}^2{\mathrm{q}}^4=9{\cdot }^6{\mathrm{C}}_4{\mathrm{p}}^4{\mathrm{q}}^2\\ \Rightarrow 9{\mathrm{p}}^2={\mathrm{q}}^2\\ \Rightarrow \mathrm{p}=\frac{1}{3}\mathrm{q}\\ \text{ We know that, }\mathrm{p}+\mathrm{q}=1\end{array} \end{gathered}$$
$\begin{array}{l}\Rightarrow \frac{\mathrm{q}}{3}+\mathrm{q}=1\\ \Rightarrow \mathrm{q}=\frac{3}{4}\\ \text{ And }\mathrm{p}=\frac{1}{4}\\ \therefore \text{ Variance }=\mathrm{npq}\\ =6\cdot \frac{1}{4}\cdot \frac{3}{4}=\frac{9}{8}\end{array}$