If X follows binomial distribution with parameters n=9 and p such that P(X=4)=4P(X=5). Then P(X=0) is :

$\left(\frac{4}{5}\right)^{9}$

The correct answer is Option (4) → $\left(\frac{4}{5}\right)^{9}$

Given that X follows a Binomial Distribution with parameters $n=9$ and $p$.

$X∼Bin(9,p)$

$P(X=k)={^9C}_k{p^k}(1-p)^{9-k}$

and,

$P(X=4)=4(X=5)$

${^9C}_4p^4(1-p)^5=4×{^9C}_5p^5(1-p)^4$

$⇒p^4(1-p)^5=4p^5(1-p)^4$

$⇒1-p=4p$

$⇒p=\frac{1}{5}$

$∴P(X=0)={^9C}_0(\frac{1}{5})^0(\frac{4}{5})^9=(\frac{4}{5})^9$