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Binomial vs Poisson distribution

Both distributions model counts of discrete events. Choosing the wrong one on the CQE exam costs you the answer even if your arithmetic is perfect.

	Binomial	Poisson
What you count	Defective units in a sample of n	Defects (events) per unit, area, or time
Fixed n?	Yes—sample size is fixed	No fixed n; n is effectively very large
p (probability)	Fixed and known for each trial	Small; λ = np is what you use
P(X=k)	C(n,k) × pᵏ × (1−p)ⁿ⁻ᵏ	e⁻λ × λᵏ / k!
Mean	np	λ
Variance	np(1−p)	λ

Use Poisson as an approximation to the binomial when n ≥ 20 and p ≤ 0.05 (or n ≥ 100 and p ≤ 0.10). Set λ = np.

Worked example 1 — Binomial

A process has 5% defective parts. A sample of n = 10 is drawn. Find P(X = 0) and P(X ≥ 1).

P(X = 0) = C(10,0) × (0.05)⁰ × (0.95)^10 = 1 × 1 × (0.95)^10 = (0.95)^10 = 0.5987 P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.5987 = 0.4013

Worked example 2 — Poisson

Surface defects occur at a rate of λ = 3 per panel. Find P(X = 0) and P(X = 2).

P(X = 0) = e^(−3) × 3⁰ / 0! = e^(−3) = 0.0498 P(X = 2) = e^(−3) × 3² / 2! = e^(−3) × 9/2 = e^(−3) × 9/2 = 0.2240

Practice problem

A 200-square-metre roll of fabric has an average of 0.5 blemishes per square metre. What is the probability of zero blemishes in a 1 m² section? (λ = 0.5)

Show answer

P(X=0) = e⁻⁰·⁵ = 0.6065. About 61% of sections will be blemish-free.

Common exam traps

“Defective units” vs “defects per unit.” Counting which units fail = binomial. Counting how many scratches are on a unit = Poisson.
Poisson requires independence. Defects must occur independently at a constant average rate. Clustered defects (one scratch causes another) violate this assumption.
np for the Poisson approximation must be small. The approximation is poor when p is large (e.g., 20% defective). Stick with the exact binomial formula in that case.

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