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, $P(X=4)={ }^{4} C_{4} p^{4} q^{0}=p^{4}$
$=\left(\frac{1}{3}\right)^{4}=\frac{1}{81}$