Binomial Distribution

Count successes in a fixed number of trials

A binomial variable X counts successes in n trials. Each trial has two outcomes, the success probability p is the same across trials, and outcomes are independent. “Success” is a label for the event being counted; it need not be desirable.

These conditions matter. Ten pageviews from one person are not necessarily ten independent purchase opportunities. Sampling without replacement from a small finite population usually changes success probabilities and calls for a hypergeometric model instead.

Worked example: four hypothetical inspections

Assume each of four independently produced items has a 20% defect probability. With X counting defects, X ~ Binomial(4, 0.2). This is an illustrative model, not an empirical manufacturing claim.

$P(X=k)=\binom nk p^k(1-p)^{n-k}.$

For exactly one defect, choose which of the four items is defective and multiply the probabilities of its outcome pattern:

$P(X=1)=4(0.2)(0.8)^3=\mathbf{0.4096}.$

Defects k	Exact probability P(X = k)	Cumulative P(X ≤ k)
0	0.4096	0.4096
1	0.4096	0.8192
2	0.1536	0.9728
3	0.0256	0.9984
4	0.0016	1.0000

The exact probabilities sum to 1. “At most one” is 0.8192, while “at least one” is 1 − P(X = 0) = 0.5904. For an upper tail including k, subtract P(X ≤ k − 1), not P(X ≤ k).

Expected count and variation

$E[X]=np,\qquad \operatorname{Var}(X)=np(1-p).$

Here the expected count is 0.8, the variance 0.64, and the standard deviation 0.8 defects. A count on a single inspection batch cannot equal 0.8; the expectation describes a long-run average under the model.

For the observed proportion X/n, the model standard deviation is √[p(1 − p)/n]. Estimating p from data introduces estimation uncertainty; an observed defect rate should not be treated as a known constant without qualification.

Decide whether the binomial assumptions are plausible

Ask whether the number of trials was fixed in advance, success was defined consistently, shared conditions create dependence, and p changes across trials. A production fault that affects an entire batch can produce more variation than the binomial model predicts.

Normal approximations can be useful when both np and n(1 − p) are sufficiently large for the desired tail accuracy, but no single cutoff guarantees accuracy everywhere. With n = 4 and p = 0.2, exact calculations are straightforward and preferable.

Practice

For three independent trials with p = 0.5, find the probability of at least two successes.

Show the explanation

P(X ≥ 2) = C(3,2)(0.5)³ + C(3,3)(0.5)³ = 3/8 + 1/8 = 0.5. The outcome patterns, not just the two possible counts, determine the weights.

Continue

Use the binomial reference, or compare discrete models. NIST gives the binomial probability functions and moments.