Poisson Approximation to the Binomial
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Direct answer
Rare events over many trials outgrow the binomial formula's arithmetic, and the Poisson distribution takes over as its limit. When n grows large, p shrinks small, and the product λ = np stays moderate (rules of thumb: n ≥ 50 with p ≤ 0.1, and λ under about 10), the binomial probability P(X = k) = C(n,k) p^k (1-p)^(n-k) is well approximated by e^(-λ) λ^k / k!. The approximation inherits the binomial's mean: the Poisson parameter λ equals np, and the Poisson variance also equals λ — the signature property distinguishing it from the binomial, whose variance npq falls below its mean. Historically this limit modelled the classical rare-event data set: Bortkiewicz's 1898 study of Prussian cavalry deaths by horse kick, where deaths per corps-year followed the Poisson law almost exactly.
What you must remember
- The approximation: P(X = k) ≈ e^(-λ) λ^k / k! with λ = np, used when n is large and p is small; common thresholds quoted are n ≥ 50, p ≤ 0.1, λ = np ≤ 10.
- Mean and variance: the approximating Poisson has mean λ and variance λ (variance = mean); the binomial it replaces has mean np and variance npq, so the gap between mean and variance signals which model produced the data.
- Why it works: (1 - p)^(n-k) ≈ e^(-np) for small p, and C(n,k) p^k → λ^k / k! as n → ∞ with np fixed — the limit theorem behind the plug-in formula.
- Additivity: sums of independent Poisson variables are Poisson with parameter the sum of λ's — two independent Poisson streams (rates 2 and 3) merge into rate 5.
- Sequential Poisson events: if events occur independently at rate λ per unit time, the count in time t is Poisson with parameter λt — the bridge to radioactivity and queuing questions.
- Anchor numbers: e^(-1) ≈ 0.3679, e^(-2) ≈ 0.1353, e^(-3) ≈ 0.0498 — JEE numerical answers live near these.
- Zero-event probability: P(X = 0) = e^(-λ) is both the easiest Poisson value and the one most tested — "probability of no defect/no call/no decay".
Measuring the approximation's quality
A factory ships lots of n = 100 items with defect probability p = 0.03 per item, so λ = np = 3. The probability of a clean lot is exactly 0.97^100; taking logarithms, 100 ln(0.97) ≈ 100 × (-0.03046) = -3.046, so 0.97^100 ≈ e^(-3.046) ≈ 0.0476. The Poisson estimate is e^(-3) ≈ 0.0498 — agreement within about 4.6 percent, on the cautious side, and for a single-decimal answer both round to 0.05. Extend to k = 1: exact, 100 × 0.03 × 0.97^99 ≈ 3 × 0.0491 ≈ 0.147; Poisson, 3e^(-3) ≈ 0.149. The lesson generalises — the approximation degrades as p climbs, which is precisely why the p ≤ 0.1 guidance exists, and it improves as n grows with λ fixed. When a question instead fixes n and p both moderate, drop the approximation and compute binomial terms directly; the examiner's phrase "rare" or "large number of trials" is the cue for λ = np.
How the exam frames it
JEE Main asks this as a formula evaluation: n = 400, p = 0.01, find P(X = 0) or P(X = 1), answers like e^(-4) or 4e^(-4); also mean-variance identification (mean 5, variance 5 identifies Poisson; mean 5, variance 4 identifies binomial with q = 0.8). Advanced dresses the same limit in word problems — misprints per page, phone calls per hour at λt, radioactive decays — where the examinee must first decide binomial versus Poisson, then often use additivity of independent Poisson counts. The characteristic errors: computing λ as n + p or np²; using λ = np but forgetting the e^(-λ) factor (answers inflated by e^λ, an order of magnitude); and applying the approximation with p = 0.4 because n is large — large n alone does not license it. One recurring true/false: Poisson as limit of binomial requires p → 0 with np fixed, not merely n → ∞. The topic is explicit probability-distributions syllabus for Main and appears in Advanced within modelling contexts.
Frequently asked questions
When may the binomial be approximated by a Poisson distribution?
When n is large and p is small with λ = np moderate — the working guidelines are n ≥ 50 with p ≤ 0.1 and λ below about 10.
What parameter does the approximating Poisson use?
λ = np, the binomial's mean; the Poisson then assigns P(X = k) ≈ e^(-λ) λ^k / k!.
How do the mean and variance distinguish the two distributions?
Poisson has variance equal to the mean (both λ); binomial has variance npq, strictly below its mean np — a data set with variance exceeding its mean fits neither.
What is the probability of zero events under the Poisson law?
P(X = 0) = e^(-λ), the most frequently examined value; for λ = 3 it is e^(-3) ≈ 0.0498.
What happens when two independent Poisson counts are added?
The sum is Poisson with parameter λ1 + λ2 — additivity, which lets merged streams (two machines, two phone lines) be treated as one process.