Question

If in a binomial distribution $n=4,P(X=0)=\frac{16}{81}$, then $P(X=4)$ equals

• A. $\frac{1}{16}$
• B. $\frac{1}{81}$
• C. $\frac{1}{27}$
• D. $\frac{1}{8}$

Answer 1

Answer: (B)

Let $p$ be the probability of success and $q$ that of failure in trial Then,
$P(X=0)={}^4{C}_0{p}^0{q}^4=\frac{16}{81}$
$\Rightarrow q^4={\left(\frac{2}{3}\right)}^4$
$\left(\because {}^n{C}_0=1\right)$
$\Rightarrow q=\frac{2}{3}$
$\Rightarrow p=1-\frac{2}{3}=\frac{1}{3}$
Therefore, $P(X=4)={}^4{C}_4{p}^4{q}^0=p^4$
$={\left(\frac{1}{3}\right)}^4=\frac{1}{81}$