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

$\frac{1}{81}$

We have,

$n=4, P(X=0)=\frac{16}{81}$

Let p be the probability of success and q that of failure in a trial.

Then,

$ P(X=0)=\frac{16}{81}$

$⇒ {^4C}_0p^0q^4=\frac{16}{81}$

$⇒q^4=\left(\frac{2}{3}\right)^4 ⇒q= \frac{2}{3}⇒ p = \frac{1}{3}$

$∴ P(X=4)= {^4C}_4 p^4q^0=p^4 = \left(\frac{1}{3}\right)^4=\frac{1}{81}$