Suppose X ~ B(n, p) and P(X = 3) = P(X = 5). If $\mathrm{p}>\frac{1}{2}$then

$\mathrm{P}(\mathrm{x}=3)=\mathrm{P}(\mathrm{x}=5),\mathrm{P}>\frac{1}{2}\Rightarrow 5\le \mathrm{n}\le 7$

${ }^n{C}_3{p}^{n-3}{q}^3{=}^n{C}_5{p}^{n-5}{q}^5$

$\frac{n!}{\left(n-3\right)!3!}{p}^2=\frac{n!}{\left(n-5\right)!5!}{q}^2$

$\frac{\left(n-3\right)\left(n-4\right)}{20}={\left(\frac{p}{1-p}\right)}^2>1$

$$\begin{gathered} n^2-7n+12>20 \\ \Rightarrow n^2-7n-8>0 \end{gathered}$$

$\left(n-8\right)\left(n+1\right)>0\Rightarrow n>8$