The Binomial Distribution

Suppose $n$ independent Bernoulli trials are performed, each with probability $p$ of success and $q = 1-p$ of failure. Let $X$ be the random variable counting the total number of successes in the $n$ trials; $X$ can take any of the values $0, 1, 2, \dots, n$. What is the probability of getting exactly $x$ successes?

Building the formula. Fix a particular arrangement of the $n$ trials with exactly $x$ successes and $n-x$ failures — say, the first $x$ trials succeed and the remaining $n-x$ fail. Because the trials are independent, the probability of this one specific sequence is found by multiplying the individual probabilities together:

$\underbrace{p \times p \times \cdots \times p}_{x \text{ times}} \times \underbrace{q \times q \times \cdots \times q}_{(n-x) \text{ times}} = p^x q^{\,n-x}$.

But this is only one of the many orders in which $x$ successes and $n-x$ failures can occur among $n$ trials — the successes could fall on any $x$ of the $n$ trial positions, and by the counting rule for combinations there are exactly $^{n}C_x$ such arrangements. Since these arrangements are mutually exclusive alternative ways of getting $x$ successes, their probabilities add up. Hence:

$P(X = x) = {}^{n}C_x \, p^x q^{\,n-x}, \qquad x = 0, 1, 2, \dots, n$.

A discrete random variable following this rule is said to have a Binomial distribution with parameters $n$ and $p$, written $X \sim B(n, p)$. It applies exactly when the four Bernoulli-trial conditions of the previous section hold: fixed $n$, two outcomes per trial, independence, and constant $p$.

Why "binomial". Notice that $^{n}C_0\,q^n, \; {}^{n}C_1\,p\,q^{n-1}, \; {}^{n}C_2\,p^2q^{n-2}, \dots, {}^{n}C_n\,p^n$ are exactly the successive terms in the binomial expansion of $(q+p)^n$. That is why $\sum_{x=0}^{n} P(X=x) = (q+p)^n = 1^n = 1$ automatically — the probabilities are guaranteed to add to $1$ simply because $p + q = 1$, without any extra checking. …