If X follows binomial distribution with parameters n=9 and p such that P(X=4)=4P(X=5). Then P(X=0) is :
Answer & explanation
Correct answer: option 4
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,
$P(X=4)=4(X=5)$
${^9C}_4p^4(1-p)^5=4×{^9C}_5p^5(1-p)^4$
$⇒p^4(1-p)^5=4p^5(1-p)^4$
$⇒1-p=4p$
$⇒p=\frac{1}{5}$
$∴P(X=0)={^9C}_0(\frac{1}{5})^0(\frac{4}{5})^9=(\frac{4}{5})^9$