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

$P(x=3)=P(x=5)andP>\frac{1}{2}$

$\Rightarrow {}^{\mathrm{n}}\mathrm{c}_3 {\mathrm{p}}^3{\mathrm{q}}^{n-3}={}^{\mathrm{n}}\mathrm{c}_5{\mathrm{p}}^5{\mathrm{q}}^{\mathrm{n}-5}$

$\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{3\times 2\times 1}\times p^3(1-p{)}^{\mathrm{n}-3}=\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)(\mathrm{n}-4)}{5\times 4\times 3\times 2\times 1} p^5(1-p{)}^{\mathrm{n}-5}$

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

$\frac{\left(n-3\right)\left(n-4\right)}{20}<1$

$5\le n\le 7$