Binomial Probability Distribution

Consider a system that can only exhibit two possible outcomes. Let us label the outcomes $P$ and $Q$, and let $p$ and $q$ be the respective probabilities of these outcomes. It follows from Equation (5.3) that

$$p + q = 1.$$ (5.9)

Consider an ensemble of $N$ two-outcome systems like the one just discussed. Let $n$ be the number of systems in the ensemble that exhibit outcome $P$, and let $n'$ be the number of systems that exhibit outcome $Q$. It is evident that

$$n+ n' = N.$$ (5.10)

Let us determine the probability, $P_N(n)$, that $n$ systems in our ensemble exhibit outcome $P$. Making use of a straightforward extension of Equation (5.8), the probability that $n$ systems in the ensemble exhibit outcome $P$, and that $n'$ exhibit outcome $Q$, is

$$\underbrace{p\,p\,p\,p\cdots p}_{n}\,\underbrace{q\,q\,q\,q\,q\cdots q}_{n'} = p^n\,q^{n'}.$$ (5.11)

However, a situation in which $n$ systems in the ensemble exhibit the outcome $P$ can be achieved in many alternative ways. Let $C_N(n)$ be the number of distinct configurations of $N$ systems by which $n$ of these systems exhibit outcome $P$. Making use of a straightforward extension of Equation (5.4), as well as Equations (5.10) and (5.11), we deduce that

$$P(n) = C_N(n)\,p^n\,q^{N-n}.$$ (5.12)

Consider $n$ systems exhibiting the outcome $P$. The number of ways that these systems can be distributed between $N$ systems is

$$N\,(N-1)\,(N-2)\cdots [N-(n-1)] = \frac{N!}{(N-n)!}.$$ (5.13)

This follows, by induction from Equation (5.8), because we can choose any one of the $N$ systems to exhibit the first outcome $P$, then we can choose any one of the remaining $N-1$ systems to exhibit the second outcome $P$, and so on. However, some of the $N!/(N-n)!$ distributions will just be permutations of the systems exhibiting outcome $P$ among themselves. Such permutations do not correspond to distinct distributions. Now, the number of permutations of $n$ quantities among $n$ places is $n!$. Hence, we deduce that

$$C_N(n) = \frac{N!}{n!\,(N-n)!}.$$ (5.14)

It follows from Equation (5.12) that

$$P_N(n)= \frac{N!}{n!\,(N-n)!}\,p^n\,q^{N-n}.$$ (5.15)

The well-known algebraic expansion of a binomial of the form $(p+q)^N$ is

$$(p+q)^N =\sum_{n=1,N} \frac{N!}{n!\,(N-n)!}\,p^n\,q^{N-n}.$$ (5.16)

For this reason, the probability distribution (5.15) is known as the binomial probability distribution. From Equation (5.9), $p+q=1$, which implies that $(p+q)^N=1$. Thus, the previous two equations yield

$$\sum_{n=0,N} P_N(n) = 1,$$ (5.17)

in accordance with Equation (5.3).

Suppose that outcomes $P$ and $Q$ represent steps to the right and steps to the left taken by a drunken man. The net number of steps to the right taken is $m=n-n' = 2\,n-N$, where use has been made of Equation (5.10). Thus,

$$n = \frac{N+m}{2},$$ (5.18)

which implies that the probability, $P_N'(m)$, that $m$ assumes a certain value after $N$ steps is equal to the probability that $n$ assumes the value $(N+m)/2$. In other words,

$$P_N'(m) = P_N\left(\frac{N+m}{2}\right).$$ (5.19)

Suppose, finally, that some physical system $A$ can exhibit many possible outcomes, $r$, $s$, $t$, et cetera. If we are only interested in outcome $r$ then we could label all of the other outcomes `not $r$' or $\bar{r}$. In this case, we have recovered a system to which the binomial probability distribution applies.