When the Poisson Approximates a Binomial

Why Approximate a Binomial Count?

A binomial model can describe the number of successes in a fixed number of independent trials when every trial has the same success probability. When there are many trials but success is rare, calculating binomial probabilities may involve large combinations and powers. A Poisson distribution can provide a simpler approximation, using a single parameter: the expected number of successes.

The approximation is useful only if its conditions make sense for the situation. A large number of trials alone is not enough, and matching the expected count alone is not enough. The probability of success on each trial should also be small, and the original binomial assumptions must be plausible.

Definition: Suppose $X$ is a binomial count with $n$ independent trials and the same success probability $p$ on each trial, so $X\sim\operatorname{Bin}(n,p)$. When $n$ is large and $p$ is small, the Poisson approximation replaces this distribution by a Poisson distribution with parameter $\lambda=np$: $$ X\sim\operatorname{Bin}(n,p)\quad\approx\quad Y\sim\operatorname{Poisson}(\lambda),\qquad \lambda=np. $$ The approximation estimates probabilities for $X$ using the corresponding probabilities for $Y$. It is not an identity: $X$ remains a binomial count in the original model.

For a Poisson variable $Y$ with parameter $\lambda$, the probability of exactly $k$ events is

P(Y=k)=\frac{e^{-\lambda}\lambda^k}{k!},\qquad k=0,1,2,\ldots.

Here $\lambda$ is the expected count in the same set of trials or observation window. In this approximation it is calculated from the binomial inputs, $n$ and $p$; it is not an additional quantity to choose independently.

What to Check Before Using the Approximation

First check the binomial model itself. The count must refer to a fixed number $n$ of trials, each trial must have two outcomes for the purpose of the count (success or not), trials must be independent, and the probability $p$ of success must be the same on every trial. If these conditions fail, simply substituting a value of $np$ into a Poisson formula does not fix the model.

Next consider whether successes are rare relative to the number of trials. The approximation is usually reasonable when $n$ is large and $p$ is small. These are qualitative conditions, not a universal pass-or-fail cutoff. Some introductory rules of thumb use $n\geq 20$ and $p\leq 0.05$, but those numbers are guides, not guarantees. A value of $p$ just beyond a guideline does not automatically make an approximation useless; a value within it does not certify accuracy for every purpose.

Finally calculate $\lambda=np$. This parameter gives the expected number of successes under the binomial model, and the Poisson model has the same mean. It also determines the spread of the approximating distribution. A useful supporting check is that the binomial variance $np(1-p)$ is close to $np$ when $p$ is small. This similarity supports the approximation, but it does not establish that every probability is close.

Verify the binomial setup.
Identify the fixed number of trials, the success definition, independence, and a common success probability.

Assess rarity.
Ask whether $n$ is large and $p$ is small in context. Treat numerical rules of thumb as guidance rather than proof.

Compute the Poisson parameter.
Calculate $\lambda=np$, retaining the units or context of the count.

Approximate the requested probability.
Use the Poisson probability formula or the appropriate sum or complement, and describe the answer as an approximation for the original binomial count.

Worked Applications

Worked Example: At Least One Rare Event

Suppose, hypothetically, a laboratory inspects 500 independent samples. Each sample has probability $0.004$ of showing a specified feature. Let $X$ be the number of samples showing the feature. Estimate the probability that at least one of the 500 samples shows it.

Check the setup. There are a fixed 500 trials, each sample either shows the feature or does not, the trials are assumed independent, and the probability $p=0.004$ is the same for every sample. Thus a binomial model is appropriate under the stated assumptions. The trial count is large and the success probability is small, so a Poisson approximation is reasonable to use.

Find the parameter.

\lambda=np=500(0.004)=2.

“At least one” is the complement of zero successes. Using the Poisson model,

P(X\geq 1)\approx P(Y\geq 1) =1-P(Y=0) =1-\frac{e^{-2}2^0}{0!} =1-e^{-2} \approx 1-0.1353 =0.8647.

The estimated probability is about $0.8647$, or $86.47\%$. In context, under the stated model assumptions, the chance of at least one sample showing the feature is approximately $86.47\%$. This is an estimate, not an exact binomial probability.

Worked Example: An Exact Count

Imagine that 120 independent components are checked, each with probability $0.02$ of having a specified defect. Let $X$ count the defective components. Estimate the probability that exactly three components are defective.

Check the setup. The number of components is fixed, each component is classified as defective or not, and the model assumes independence and a common defect probability of $0.02$. With 120 trials and a small probability of success, the Poisson approximation is reasonable as a working estimate.

The parameter is

\lambda=np=120(0.02)=2.4.

Substitute $k=3$ and $\lambda=2.4$ into the Poisson probability mass function:

P(X=3)\approx P(Y=3) =\frac{e^{-2.4}(2.4)^3}{3!} =\frac{(0.09072)(13.824)}{6} \approx 0.2090.

The estimated probability of exactly three defective components is about $0.2090$, or $20.90\%$. The parameter $2.4$ is the expected number of defects in the 120 components, not the probability of a defect on one component.

Worked Example: When the Success Probability Is Not Small

Suppose, hypothetically, that 20 independent units each have probability $0.20$ of passing a particular screening criterion. A proposed analysis uses a Poisson approximation to estimate the probability of exactly four passes. Decide whether that approximation is well supported by the usual conditions, and identify the parameter it would use.

Check the setup. The situation has a fixed number of trials, two outcomes for each unit, and—if the stated assumptions hold—independent trials with a common success probability. A binomial model is therefore plausible.

The proposed Poisson parameter would be

\lambda=np=20(0.20)=4.

Although $n=20$ is not especially small, $p=0.20$ is not a small success probability. The condition that makes individual successes rare is not well met. The value $\lambda=4$ does not resolve that concern: it matches the expected count but does not ensure that the binomial probabilities are close to Poisson probabilities. Therefore, the approximation is not well supported here; the binomial model should be retained for the probability calculation. This decision concerns the suitability of the approximation, not whether the Poisson formula can be evaluated.

Why the Conditions Matter

The approximation has a useful intuition. When $p$ is small, a success on any one trial is rare. If there are many such trials, the total count can still have a non-negligible expected value, because the product $np$ may be moderate. The Poisson model uses that product to represent the count. The binomial mean is $np$, matching the Poisson mean $\lambda$, and the binomial variance $np(1-p)$ approaches $np$ as $p$ becomes small:

np(1-p)=np-np^2\quad\text{and}\quad \frac{np(1-p)}{np}=1-p\quad (np>0).

Thus, as $p$ decreases, the binomial variance becomes closer to its mean, as in a Poisson distribution. This is a brief justification for why the Poisson model is plausible in rare-event settings. It is not a derivation of the approximation and does not guarantee that every probability, especially a probability in an extreme tail, is close. For consequential calculations, a guideline is not a substitute for checking the accuracy needed in the application.

The model assumptions also matter independently of the approximation. If one event changes the chance of another, or if different trials have substantially different success probabilities, the binomial starting model may be inappropriate. Replacing it with a Poisson distribution does not remove dependence or make unequal probabilities equal. The context should support the assumptions before the approximation is considered.

Common Mistake

Wrong reasoning: “The mean count is $np=4$, so the Poisson approximation is suitable.” This uses only the matched mean and ignores both the size of $p$ and the binomial assumptions.

Correction: Calculate $np$, but also check that the number of trials is large, each success probability is small, and the fixed-trial, independence, and equal-probability conditions are credible. A suitable value of $\lambda$ alone is not evidence that the approximation is accurate.

Key takeaway: Approximate $\operatorname{Bin}(n,p)$ by a Poisson distribution with $\lambda=np$ when the binomial assumptions are credible, $n$ is large, and $p$ is small. State the result as an approximation and keep the original binomial model in view.

Check Your Understanding

For each question, explain the reasoning rather than relying only on a numerical rule.

A fixed set of 800 independent items each has probability $0.003$ of a specified outcome. What Poisson parameter would you use, and what features support the approximation?
A binomial count has $n=30$ and $p=0.15$. Why might calculating $\lambda=np$ be insufficient to justify a Poisson approximation?
In a binomial setting with $n=250$ and $p=0.01$, estimate the probability of no successes using a Poisson model. Show the parameter and the probability calculation.
Why does the Poisson approximation not fix a setting in which the trials are dependent?
Explain why matching the binomial mean to the Poisson mean supports the approximation but does not guarantee accuracy for every probability.