Why Model Choice Matters for Counts
A count problem may describe the number of claims, defects, late arrivals, or other events in a fixed group or time period. The calculation depends on how that count is generated. A fixed number of independent opportunities, each with the same event probability, leads to a binomial model. A Poisson model may be appropriate for events occurring at a stable average rate, or as an approximation to a binomial count when individual event probabilities are small. A normal approximation can be useful when a binomial count has enough expected successes and failures to have a roughly symmetric distribution.
Choosing a convenient formula is not enough. First identify the data-generating assumptions; then assess whether an approximation is credible for the probability being asked about. In particular, an approximation can be adequate for a broad central range but less reliable for a very small tail probability.
Write \(q=1-p\) when it is useful to shorten expressions. The binomial probability of exactly \(k\) counted outcomes is
A binomial model is not justified merely because the quantity being counted is a whole number. If trial probabilities vary, or outcomes are dependent, the binomial assumptions may fail. A fixed population sampled without replacement, for example, generally creates dependence between selections.
Three Choices and Their Conditions
| Choice | When it is appropriate | Parameters or calculation |
|---|---|---|
| Exact binomial | Fixed number of independent trials, common event probability | \(n\) and \(p\); use the binomial pmf or add relevant probabilities |
| Poisson model or approximation | Events occur independently at a stable rate, or a binomial has large \(n\) and small \(p\) | For a binomial approximation, \(\lambda=np\); use \(P(X=k)\approx e^{-\lambda}\lambda^k/k!\) |
| Normal approximation | A binomial count has sufficiently many expected successes and failures | Approximate with \(N(np,npq)\), using the binomial mean and variance |
The Poisson distribution may be the model itself: for instance, a count over a specified interval may be modeled as Poisson with a stated rate parameter. In that case, its assumptions must be reasonable for the situation. Alternatively, it may approximate a binomial count. In the latter case, use \(\lambda=np\), not \(p\) alone.
Common rules of thumb say that a binomial-to-Poisson approximation is most promising when \(n\) is large and \(p\) is small, and that a normal approximation is often considered when both \(np\) and \(nq\) are at least 5 or 10. These are screening guidelines, not guarantees. No single cutoff makes every tail probability accurate. When the binomial sum is manageable, compare an approximation with the exact calculation.
Why the Poisson Approximation Arises
The binomial-to-Poisson formula follows by considering \(n\) large, \(p\) small, and the product \(np=\lambda\) held fixed. For a fixed nonnegative integer \(k\), set \(p=\lambda/n\) in the binomial pmf:
Rewrite the factors so their limits are visible:
As \(n\) increases with \(k\) fixed, the bracketed product tends to 1, the middle power tends to \(e^{-\lambda}\), and the last factor tends to 1. Therefore the limiting probability is \(e^{-\lambda}\lambda^k/k!\), the Poisson pmf. This explains why the approximation can work for rare events across many trials. It does not imply that every finite-\(n\) binomial tail is close to its Poisson counterpart.
Worked Examples: Select and Apply a Count Model
Worked Example: Use the Exact Binomial Model When Trials Are Few
A hypothetical inspection system checks 10 independently produced components. Each component has probability \(0.20\) of failing inspection, independently of the others. Find the probability that at most one component fails.
Method and model. The number of components is fixed at \(n=10\), the outcomes are failure or no failure, and each component has the same failure probability \(p=0.20\). Thus \(X\sim\operatorname{Bin}(10,0.20)\), with \(q=0.80\). Because \(n\) is modest and \(p\) is not especially small, use the exact binomial model.
“At most one” means \(X=0\) or \(X=1\), so
Calculate each term:
Adding gives \(0.1073741824+0.268435456=0.3758096384\). Rounded to four decimal places, the probability is 0.3758. As a check, the two disjoint cases cover exactly the requested outcomes, zero and one failures.
A Poisson approximation would use \(\lambda=np=2\), giving \(e^{-2}(1+2)\approx0.4060\), which differs from the exact result by about \(0.0302\). This example shows why “a count” alone is not a reason to approximate: the exact calculation is straightforward, and the trials are not in a particularly rare-event regime.
Worked Example: Approximate a Rare-Event Binomial Count by Poisson
A hypothetical monitoring program examines 100 independent devices. Each has probability \(0.02\) of producing a specified fault during the observation period. Estimate the probability of at most one fault, and compare the Poisson approximation with the exact binomial result.
Method and model. The exact count has \(X\sim\operatorname{Bin}(100,0.02)\). Since \(n=100\) is relatively large and \(p=0.02\) is small, a Poisson approximation is reasonable to check. Its parameter is the expected count \(\lambda=np=100(0.02)=2\).
For a Poisson random variable \(W\) with parameter 2,
Now calculate the exact binomial result for comparison:
The first term is \(P(X=0)=(0.98)^{100}\approx0.1326196\). The second is \(P(X=1)=100(0.02)(0.98)^{99}\approx0.2706522\). Thus the exact probability is approximately \(0.1326196+0.2706522=0.4032718\), or 0.4033 to four decimal places. The Poisson estimate is higher by about \(0.0027\). For this event, that is a fairly close approximation; it is not an identity, and the exact value is available for comparison.
The check is that both calculations use the same event, at most one fault, and the Poisson parameter is the expected number of faults, 2. The method avoids treating the individual probability \(0.02\) as the Poisson parameter.
Worked Example: Use a Normal Approximation for a Broad Count Range
A hypothetical batch contains 1,000 independent sensors, each of which has probability \(0.30\) of registering a warning during a test. Estimate the probability that the number of warnings is between 270 and 330, inclusive, using a direct normal approximation.
Method and model. Let \(X\sim\operatorname{Bin}(1000,0.30)\). Here \(n=1000\), \(p=0.30\), and \(q=0.70\). The expected successes and failures are \(np=300\) and \(nq=700\), both large, so the count is sufficiently far from the strongly skewed rare-event setting for a normal approximation to be plausible.
The binomial mean and variance are
Therefore the approximating normal distribution is \(Y\sim N(300,210)\), where the second parameter is the variance. Its standard deviation is \(\sigma=\sqrt{210}\approx14.4914\). Standardizing the two stated endpoints gives
Using the standard normal cumulative distribution function \(\Phi\),
The estimated probability is 0.9616 to four decimal places. The calculation is symmetric around the mean, which provides a useful check on the equal-and-opposite standardized endpoints. This is an approximation, not an exact binomial probability; integer cutoffs and tail accuracy can affect how well a continuous approximation performs.
Common Exam Traps
- Using \(p\) instead of \(np\) as the Poisson parameter. For \(n=100\) and \(p=0.02\), the approximation parameter is \(\lambda=2\), not \(0.02\). The incorrect calculation for at most one event would be \(e^{-0.02}(1+0.02)\approx0.9998\); the correct Poisson calculation is \(e^{-2}(1+2)\approx0.4060\).
- Using the variance as the standard deviation. For a binomial count the variance is \(npq\), but standardization divides by \(\sqrt{npq}\). In the sensor example, the standard deviation is \(\sqrt{210}\approx14.4914\), not 210.
- Approximating before checking the model. A normal or Poisson formula does not repair dependence between trials or varying event probabilities. Check how the count is generated first.
- Treating a rule of thumb as a guarantee. Large expected counts support a normal approximation, and small \(p\) supports a Poisson approximation, but accuracy depends on the particular event and the desired precision.
Check Your Understanding
For each problem, choose the model or approximation indicated by the conditions and check the assumptions before calculating.
- A device has probability \(0.10\) of triggering an alarm on each of 10 independent tests. Find the exact probability of no alarms. (Answer: \(0.9^{10}\approx0.3487\))
- Among 80 independent accounts, each has probability \(0.025\) of generating a particular notice. Use a Poisson approximation to find the probability of at most two notices. (Answer: \(\lambda=2\), so \(e^{-2}(1+2+2)\approx0.6767\))
- A count is modeled as \(X\sim\operatorname{Bin}(400,0.40)\). Using a direct normal approximation, estimate \(P(X\leq170)\). (Answer: \(\mu=160\), \(\sigma=\sqrt{96}\approx9.7980\), \(z\approx1.0206\), so the estimate is \(0.8462\))
- A batch has 40 items, each independently defective with probability \(0.30\). Which model should be used for an exact probability calculation, and why is a Poisson approximation not the automatic first choice? (Answer: Use \(X\sim\operatorname{Bin}(40,0.30)\) exactly; the event probability is not especially small, and the exact model is specified by the independent fixed-trial assumptions.)
- For \(X\sim\operatorname{Bin}(100,0.01)\), compare the exact probability of zero events with its Poisson approximation. (Answer: Exact \(P(X=0)=0.99^{100}\approx0.3660\); Poisson uses \(\lambda=1\), giving \(e^{-1}\approx0.3679\).)