If X follows Binomial distribution with parameters ${\rm{n}} = 5,$ ${\rm{p}}$ and $P(X = 2) = 9P(X = 3)$, then $p$ is equal to……………

(where $,n = 5$ and $q = 1 - p$)
$\Rightarrow$ $^5{C_2}{p^2}{(1 - p)^3} = 9{ \cdot ^5}{C_3}{p^3}{(1 - p)^2}$

$\Rightarrow$ $\frac{{5!}}{{2!3!}}{p^2}{(1 - p)^3} = 9 \cdot \frac{{5!}}{{3!2!}}{p^3}{(1 - p)^2}$

$\Rightarrow$ $\frac{{{p^2}{{(1 - p)}^3}}}{{{p^3}{{(1 - p)}^2}}} = 9$

$\Rightarrow$ $\frac{{(1 - p)}}{p} = 9 \Rightarrow 9p + p = 1$

$\therefore p = \frac{1}{{10}}$