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

(b) $\frac{1}{81}$

In the given binomial distribution, n = 4 and
$$\begin{gathered} P(X=0)=\frac{16}{81} \\ \mathrm{Binomial}\mathrm{distribution}\mathrm{is}\mathrm{given}\mathrm{by} \\ P(X=0)={}^{4}C_0{p}^0{q}^{4-0}=q^4 \\ \text{We}\mathrm{know}\mathrm{that}P(X=0)=\frac{16}{81} \\ \therefore q^4=\frac{16}{81} \\ \Rightarrow q^4={\left(\frac{2}{3}\right)}^4 \\ \Rightarrow q=\frac{2}{3} \\ \therefore p=1-\frac{2}{3}=\frac{1}{3} \\ \mathrm{Then},P(X=4)={}^{4}C_4{p}^4{q}^{4-4} \\ ={\left(\frac{1}{3}\right)}^4 \\ =\frac{1}{81} \end{gathered}$$