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

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

For a binomial distribution, the mean is np and the variance is npq:

    \(np = 5,\quad npq = 4\)

Dividing the second by the first:

    \( \displaystyle q = \frac{4}{5},\quad p = 1 - q = \frac{1}{5} \)

    \( \displaystyle n = \frac{5}{p} = 25 \)

So

    \( \displaystyle P(X = r) = {}^{25}C_{r}\left(\frac{1}{5}\right)^{r}\left(\frac{4}{5}\right)^{25-r},\quad r = 0, 1, \ldots, 25 \)

Why the other options are incorrect

• Option (1) \(P(X = r) = {}^{25}C_{r}\left(\displaystyle\frac{4}{5}\right)^r\left(\displaystyle\frac{1}{5}\right)^{25-r},\,\,r = 0,1,2,3,...,25\) — Incorrect. It swaps p and q: with \(p = \frac{4}{5}\) the mean would be \(25 \times \frac{4}{5} = 20\), not 5.

• Option (3) \(P(X = r) = {}^{25}C_{r}\left(\displaystyle\frac{1}{5}\right)^{25}\left(\displaystyle\frac{4}{5}\right)^{25-r},\,\,r = 0,1,2,3,...,25\) — Incorrect. The power of p must be r, not 25.

• Option (4) \(P(X = r) = {}^{25}C_{r}\left(\displaystyle\frac{1}{5}\right)^{25-r}\left(\displaystyle\frac{4}{5}\right)^{25},\,\,r = 0,1,2,3,...,25\) — Incorrect. The powers of p and q must be r and \(25 - r\), adding up to 25.