Choosing a Model for a Count
A count such as the number of faulty components in a shipment or the number of calls received in an hour is discrete. When its distribution is known, an exact probability calculation is usually the clearest starting point. When the count comes from many trials or has a convenient approximate model, a Poisson or normal distribution can make calculation easier—but only if the model’s assumptions and approximation conditions are plausible.
This tutorial focuses on binomial and Poisson counts. For a binomial count, a Poisson approximation is most useful when the number of trials is large and the probability of success on each trial is small. A normal approximation is often useful when the expected numbers of successes and failures are both large enough. Neither rule guarantees accuracy for every event, especially in a far tail.
The binomial model requires a fixed number of trials, two outcomes per trial, independence, and a constant success probability. Use a Poisson model directly when the count is specified to follow a Poisson distribution; for example, a hypothetical system might model its number of alarms per hour that way. In either case, identify the event carefully before calculating: “at most 3” means \(X\leq3\), while “fewer than 3” means \(X\leq2\).
Approximating a Binomial Count by a Poisson Count
For \(X\sim\operatorname{Bin}(n,p)\), the Poisson approximation uses \(\lambda=np\). It is a practical fit when \(n\) is large and \(p\) is small, so that the count is driven by many opportunities, each with a low probability of success. Rules of thumb such as \(n\geq20\) and \(p\leq0.05\) are guidelines, not guarantees. The suitability of an approximation can depend on the event being evaluated.
The reason for using \(\lambda=np\) can be seen from the probability of a fixed count \(k\). Set \(p=\lambda/n\) in the binomial probability and let \(n\) grow while \(\lambda\) and \(k\) stay fixed. Then
In this limit, the factor involving the \(k\) trial positions approaches \(1/k!\), and \((1-\lambda/n)^n\) approaches \(e^{-\lambda}\). This gives the Poisson probability formula. The argument explains why the approximation is natural; it does not say that any finite \(n\) and \(p\) produce an exact Poisson distribution.
Worked Example: Approximating a Rare-Event Binomial Count
A hypothetical quality-control system checks 200 independent units. Each unit independently triggers a particular alert with probability \(0.01\). Estimate the probability that at least 3 alerts occur, first exactly and then with a Poisson approximation. Compare the two results.
Method and model. The count \(X\) is \(\operatorname{Bin}(200,0.01)\): the number of trials is fixed and the alert probability is the same on each independent trial. The event is \(X\geq3\), so calculate its complement \(P(X\leq2)\). For the Poisson approximation, \(\lambda=np=200(0.01)=2\). The small \(p\) and large \(n\) support using it as an approximation.
Exact binomial calculation. Let \(P_k=P(X=k)\). The zero-alert probability is \(P_0=(0.99)^{200}\approx0.133980\). Consecutive probabilities can be obtained from the binomial formula using \(P_{k+1}/P_k=((200-k)/(k+1))(0.01/0.99)\). Therefore,
Thus \(P(X\leq2)\approx0.133980+0.270666+0.272033=0.676679\), and \(P(X\geq3)\approx1-0.676679=0.323321\).
Poisson calculation. For \(Z\sim\operatorname{Poisson}(2)\),
The two estimates agree closely: to four decimal places, each is \(0.3233\). The exact binomial calculation is the check here; the Poisson value is not treated as exact for \(X\). Common mistake avoided: using \(\lambda=p\) rather than \(\lambda=np\) would discard the number of opportunities and give the wrong scale for the count.
Normal Approximation to a Binomial Count
For \(X\sim\operatorname{Bin}(n,p)\), the corresponding normal approximation has mean \(\mu=np\) and variance \(\sigma^2=np(1-p)\). A common practical check is that both \(np\) and \(n(1-p)\) are at least 10. This is a guideline for deciding whether the binomial shape is sufficiently bell-shaped for a useful approximation, not a theorem guaranteeing a particular error. Be especially cautious with probabilities far from the mean.
A normal variable is continuous, whereas \(X\) takes integer values. A continuity correction includes the half-unit interval around each integer count. For example, the event \(a\leq X\leq b\), where \(a\) and \(b\) are integers, is approximated by the normal probability from \(a-0.5\) to \(b+0.5\). The corresponding boundary changes for other count events are summarized below.
| Count event | Continuity-corrected normal event |
|---|---|
| \(X\leq b\) | \(Y\leq b+0.5\) |
| \(X<b\) | \(Y\leq b-0.5\) |
| \(X\geq a\) | \(Y\geq a-0.5\) |
| \(X>a\) | \(Y\geq a+0.5\) |
| \(a\leq X\leq b\) | \(a-0.5\leq Y\leq b+0.5\) |
Worked Example: A Continuity-Corrected Binomial Probability
A hypothetical monitoring program records whether each of 400 independent devices produces a specified signal during a test. Each device produces the signal with probability \(0.30\). Approximate the probability that between 110 and 130 devices, inclusive, produce the signal.
Method and checks. The count \(X\sim\operatorname{Bin}(400,0.30)\), with \(\mu=np=400(0.30)=120\) and \(\sigma^2=np(1-p)=400(0.30)(0.70)=84\), so \(\sigma=\sqrt{84}\approx9.1652\). The checks are \(np=120\geq10\) and \(n(1-p)=280\geq10\); a normal approximation is reasonable as a practical choice.
The inclusive integer event is \(110\leq X\leq130\). Applying the continuity correction gives \(109.5\leq Y\leq130.5\), for \(Y\sim N(120,84)\). Standardizing both endpoints gives
The estimated probability is \(0.7480\), to four decimal places. Common mistake avoided: using 110 and 130 directly as the normal boundaries gives approximately \(\Phi(10/9.1652)-\Phi(-10/9.1652)\approx0.725\). This omits the half-unit intervals belonging to the endpoint counts. The corrected boundaries give the appropriate approximation to the inclusive event.
Calculating an Exact Poisson Probability
A Poisson model can also be used directly rather than as an approximation to a binomial count. For \(Z\sim\operatorname{Poisson}(\lambda)\), individual probabilities are calculated from \(e^{-\lambda}\lambda^k/k!\), and probabilities over several counts are added. The recurrence \(P(Z=k+1)=P(Z=k)\lambda/(k+1)\) provides an efficient way to compute successive terms.
Worked Example: An Exact Probability for a Poisson Count
A hypothetical help desk models the number \(Z\) of overnight system notices as \(\operatorname{Poisson}(4.5)\). Calculate the probability of receiving from 3 through 6 notices, inclusive.
Method and model. The stated Poisson parameter is \(\lambda=4.5\). The event is \(3\leq Z\leq6\), so sum the four Poisson probabilities. Start with \(P(Z=0)=e^{-4.5}\approx0.0111090\), then apply the recurrence:
Add only the terms in the requested range: \(P(3\leq Z\leq6)\approx0.1687179+0.1898076+0.1708268+0.1281201=0.6574724\). The answer is \(0.6575\) to four decimal places. Common mistake avoided: the values for 0, 1, and 2 were needed to generate the recurrence, but they are not part of the event and must not be included in its final sum.
Selection Checks and Exam Traps
A reliable solution states the model before using a formula, translates the event into an inequality, and then checks whether the proposed approximation is sensible. If the binomial probabilities are manageable, an exact calculation is a useful benchmark. If an approximation is used, report it as an estimate and retain enough intermediate precision to avoid rounding error.
For a binomial count, verify the fixed number of independent trials and common success probability. For a Poisson count, state the parameter \(\lambda\) and whether the model is assumed directly or used as an approximation.
Distinguish “at most,” “fewer than,” “at least,” “more than,” and inclusive ranges. Decide whether a complement or a sum of individual probabilities is simpler.
For a Poisson approximation to a binomial count, calculate \(\lambda=np\) and assess whether \(n\) is large and \(p\) small. For a normal approximation, calculate \(np\) and \(n(1-p)\), and consider whether a tail event could be poorly estimated.
Apply a continuity correction to a normal approximation of an integer count. Check that the resulting probability lies between 0 and 1 and that a complement or sum uses exactly the requested event.
Two errors deserve particular attention. First, replacing a binomial count by a Poisson variable does not mean that every binomial with the same mean is well approximated: a large success probability can make the shape quite different. Second, a normal approximation does not remove the discrete nature of the event; omitting or reversing the half-unit correction can noticeably change a probability.
Check Your Understanding
For each problem, select the indicated exact or approximate method and show the event boundaries before calculating.
- Let \(X\sim\operatorname{Bin}(10,0.20)\). Find \(P(X=0)\) exactly. (Answer: \(0.1074\))
- A count is modeled as \(X\sim\operatorname{Bin}(100,0.02)\). Use a Poisson approximation to estimate \(P(X\leq1)\). (Answer: \(0.4060\))
- Let \(X\sim\operatorname{Bin}(100,0.50)\). Use a normal approximation with continuity correction to estimate \(P(45\leq X\leq55)\). (Answer: \(0.7287\))
- A hypothetical inspection count follows \(X\sim\operatorname{Bin}(400,0.015)\). Use a Poisson approximation to estimate the probability of at most 3 inspections resulting in a specified finding. (Answer: \(0.1512\))