The binomial distribution

It follows from Eqs. (16) and (20) that the probability of obtaining $n_1$ occurrences of the outcome 1 in $N$ statistically independent observations of a two-state system is

$$P_N(n_1) = \frac{N!}{n_1 !\,(N-n_1)!} \,p^{n_1}\,q^{N-n_1}.$$ (21)

This probability function is called the binomial distribution function. The reason for this is obvious if we tabulate the probabilities for the first few possible values of $N$ (see Tab. 1).

Table 1: The binomial probability distribution

				$n_1$
		0	1	2	3	4
	1	$q$	$p$
$N$	2	$q^2$	$2\,p\,q$	$p^2$
	3	$q^3$	$3\,p\,q^2$	$3\,p^2\,q$	$p^3$
	4	$q^4$	$4\,p\,q^3$	$6\,p^2\,q^2$	$4\,p^3\,q$	$p^4$

Of course, we immediately recognize these expressions: they appear in the standard algebraic expansions of $(p+q)$, $(p+q)^2$, $(p+q)^3$, and $(p+q)^4$, respectively. In algebra, the expansion of $(p+q)^N$ is called the binomial expansion (hence, the name given to the probability distribution function), and can be written

$$(p+q)^N \equiv \sum_{n=0}^{N} \frac{N!}{n!\,(N-n)!}\,p^n \,q^{N-n}.$$ (22)

Equations (21) and (22) can be used to establish the normalization condition for the binomial distribution function:

$$\sum_{n_1=0}^N P_N(n_1) =\sum_{n_1=0}^N \frac{N!}{n_1!\,(N-n_1)!}\,p^{n_1}\, q^{N-n_1}\equiv (p+q)^N = 1,$$ (23)

since $p+q=1$.