X is a binomial variate with parameters n = 6

and p. If $4\mathrm{P}\left(\mathrm{X}=4\right)=\mathrm{P}\left(\mathrm{X}=2\right), \mathrm{then} \mathrm{p} \mathrm{is}$

Given X follows Binomial distributions  n=6

Given that 4 P(X=4)=P(X=2)

$$\begin{gathered} \Rightarrow 4\times {}^6{\mathrm{c}}_4 {\left(\mathrm{p}\right)}^4 {\left(\mathrm{q}\right)}^{6-4}={}^6{\mathrm{c}}_2 {\left(\mathrm{p}\right)}^2 {\left(\mathrm{q}\right)}^{6-2} \\ \Rightarrow 4\times {}^6{\mathrm{c}}_4 {\mathrm{p}}^4 {\mathrm{q}}^2={}^6{\mathrm{c}}_2 {\mathrm{p}}^2 {\mathrm{q}}^4 \\ \Rightarrow 4{\mathrm{p}}^2={\mathrm{q}}^2 \\ \Rightarrow 4{\mathrm{p}}^2={(1-\mathrm{p})}^2 \Rightarrow 4{\mathrm{p}}^2=1+{\mathrm{p}}^2-2\mathrm{p} \\ \Rightarrow 3{\mathrm{p}}^2+2\mathrm{p}-1=0 \\ \Rightarrow 3{\mathrm{p}}^2+3\mathrm{p}-\mathrm{p}-1=0 \\ \Rightarrow 3\mathrm{p}(\mathrm{p}+1)-1(\mathrm{p}+1)=0 \\ (\mathrm{p}+1) (3\mathrm{p}-1)=0 \\ \Rightarrow \mathrm{p}=-1 \left(\mathrm{or}\right) \mathrm{p}=\frac{1}{3} \\ \Rightarrow \mathrm{p}=\frac{1}{3} (\because 0\le \mathrm{p}\le 1) \end{gathered}$$