The Binomial Distribution

The binomial distribution counts how many “successes” occur in a fixed number of independent yes/no trials — how many of $n$ vaccinated people avoid infection, how many of $n$ tossed coins land heads, how many of $n$ patients respond to a treatment. It is the natural model behind testing a proportion.

Figure 1. The Binomial(10, 0.3) probability mass function; the mean is np = 3.

Definition#

Let $X\sim\mathrm{Binomial}(n,p)$ count the successes in $n$ independent trials, each with success probability $p$ . Its probability mass function is $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}.$

Support: $k\in\{0,1,2,\dots,n\}$ .
Parameters: number of trials $n\in\{1,2,\dots\}$ and success probability $p\in[0,1]$ .
Mean: $\mathbb{E}[X]=np$ .
Variance: $\mathrm{Var}(X)=np(1-p)$ .

The binomial coefficient $\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$ counts the number of ways to arrange $k$ successes among $n$ trials.

Sum of Bernoulli trials#

A single trial with outcome $1$ (success, probability $p$ ) or $0$ (failure) is a Bernoulli random variable with mean $p$ and variance $p(1-p)$ . A binomial variable is just the sum of $n$ independent Bernoulli trials: $X=\sum_{i=1}^{n}B_i,\qquad B_i\sim\mathrm{Bernoulli}(p).$ Because expectation and (for independent terms) variance add, this immediately gives $\mathbb{E}[X]=np$ and $\mathrm{Var}(X)=np(1-p)$ .

Show how

np

and

np(1-p)

follow from a single trial

Write $X = B_1 + B_2 + \cdots + B_n$ as a sum of independent Bernoulli $(p)$ indicators. For one trial $\mathbb{E}[B] = p$ , and because $B$ takes only the values $0$ and $1$ we have $B^2 = B$ , so $\mathbb{E}[B^2] = p$ and

$\operatorname{Var}(B) = \mathbb{E}[B^2] - \big(\mathbb{E}[B]\big)^2 = p - p^2 = p(1-p) .$

Expectation adds over any sum, and variance adds over an independent sum, so

$\mathbb{E}[X] = \sum_{i=1}^{n}\mathbb{E}[B_i] = np, \qquad \operatorname{Var}(X) = \sum_{i=1}^{n}\operatorname{Var}(B_i) = np(1-p) .$

The identity $B^2 = B$ is the whole trick: it makes the single-trial variance $p(1-p)$ , which the $n$ independent trials simply accumulate.

When it arises#

The binomial arises whenever you count successes among a fixed number of independent, identical trials: proportion testing (fraction cured, fraction defective), survey responses (yes/no), and quality control. The sample proportion $\hat p=X/n$ is the basis for inference about an unknown $p$ .

Extension: the multinomial#

When each trial has more than two possible outcomes (say categories $1,\dots,m$ with probabilities $p_1,\dots,p_m$ summing to $1$ ), the counts follow the multinomial distribution $P(X_1=k_1,\dots,X_m=k_m)=\frac{n!}{k_1!\cdots k_m!}\,p_1^{k_1}\cdots p_m^{k_m},$ the direct generalization of the binomial to several categories.

In code#

R#

# pmf, cdf, quantile, and sampling for Binomial(n = 20, p = 0.3)
dbinom(6, size = 20, prob = 0.3)   # P(X = 6)
pbinom(6, size = 20, prob = 0.3)   # P(X <= 6)
qbinom(0.95, size = 20, prob = 0.3) # 95% quantile

set.seed(123)
x <- rbinom(10000, size = 20, prob = 0.3)  # random sample
hist(x, breaks = seq(-0.5, 20.5, 1), freq = FALSE)  # histogram
points(0:20, dbinom(0:20, 20, 0.3))                 # overlay the pmf

Python#

Python

import numpy as np
from scipy import stats

n, p = 20, 0.3
stats.binom.pmf(6, n, p)    # P(X = 6)
stats.binom.cdf(6, n, p)    # P(X <= 6)
stats.binom.ppf(0.95, n, p) # 95% quantile

rng = np.random.default_rng(123)
x = rng.binomial(n, p, size=10000)  # random sample
# plt.hist(x, bins=range(0, 22), density=True); overlay stats.binom.pmf(range(21), n, p)

Julia#

Julia

using Distributions, Random

d = Binomial(20, 0.3)   # Binomial(n, p)
pdf(d, 6)               # P(X = 6)   (pdf = pmf for discrete)
cdf(d, 6)               # P(X <= 6)
quantile(d, 0.95)       # 95% quantile

Random.seed!(123)
x = rand(d, 10_000)     # random sample
# histogram(x, normalize=:pdf); scatter!(0:20, pdf.(d, 0:20)) to overlay the pmf

Simulation#

The empirical mean of many binomial draws converges to $np$ and the variance to $np(1-p)$ .

set.seed(7)
x <- rbinom(1e6, size = 20, prob = 0.3)
mean(x)  # ~ 6.0   (theoretical mean = np = 20 * 0.3)
var(x)   # ~ 4.2   (theoretical variance = np(1-p) = 20 * 0.3 * 0.7)

Why it matters for statistics#

The binomial underpins inference for proportions: confidence intervals for $p$ , tests comparing two proportions, and the reasoning behind p-values in exact tests. For large $n$ it is well approximated by the normal distribution, and for large $n$ with small $p$ it approaches the Poisson distribution.