The binomial distribution for which the mean is 5 and variance 4, is

$P(X = r) = {^{25}C}_r\left(\frac{1}{5}\right)^r\left(\frac{4}{5}\right)^{25-r},\,\,r = 0,1,2,3,...,25$

The correct answer is Option (2) → $P(X = r) = {^{25}C}_r\left(\frac{1}{5}\right)^r\left(\frac{4}{5}\right)^{25-r},\,\,r = 0,1,2,3,...,25$ **

For a binomial distribution:

Mean $=np=5$

Variance $=npq=4$

So, $\frac{npq}{np}=q=\frac{4}{5}$ and $p=\frac{1}{5}$.

From $np=5$:

$n\left(\frac{1}{5}\right)=5 \Rightarrow n=25$

Thus the binomial distribution is:

$P(X=r)=\frac{25!}{r!(25-r)!}\left(\frac{1}{5}\right)^{r}\left(\frac{4}{5}\right)^{25-r}$

Correct answer:

$P(X=r)=\;{}^{25}C_{r}\left(\frac{1}{5}\right)^{r}\left(\frac{4}{5}\right)^{25-r}$