Choosing a Calculation for the Event You Need
A Poisson approximation can simplify binomial calculations, especially when there are many independent trials and each has a small success probability. But “many trials” and “small probability” do not by themselves guarantee that an approximation is accurate enough for a particular question. The relevant comparison is between the exact and approximate probabilities of the event being asked about, judged against the accuracy the problem requires.
For example, an approximation may be adequate for the probability of exactly two successes but not for the probability of no successes under the same model. A tolerance also matters: an absolute error of 0.004 may be acceptable if the permitted error is 0.01, but not if the permitted error is 0.001. When the exact binomial calculation is manageable, it provides a direct result and avoids relying on an approximation whose accuracy has not been established for the event.
The binomial model requires a fixed number \(n\) of trials, independent outcomes, and the same success probability \(p\) on each trial. The Poisson approximation is often most useful when \(p\) is small and \(n\) is sufficiently large that \(np\) remains a useful finite parameter. These are practical clues, not an event-specific guarantee of accuracy. Earlier material on Poisson approximation error provides general bounds; here the emphasis is the decision for a stated probability and tolerance.
A Practical Decision Rule
First write down the exact binomial event and the corresponding Poisson event. “Exactly \(k\)” means \(P(X=k)\); “at most \(r\)” means a sum from zero through \(r\); “at least one” means \(1-P(X=0)\). Then decide what accuracy is required. If the problem gives an absolute-error tolerance \(\varepsilon\), compare \[ \left|P(\text{event under the binomial model})- P(\text{same event under the Poisson model})\right| \] with \(\varepsilon\). If the difference is larger than the tolerance, the approximation does not meet the requirement. If the exact answer is accessible, use it. If it is smaller, the approximation meets that stated absolute-error criterion.
The criterion is about the event, not just the distribution in general. Relative error may also matter when a probability is very small: an absolute error that looks small can be large compared with the probability itself. Unless the question specifies otherwise, do not substitute a relative-error requirement for an absolute-error requirement, or vice versa.
Identify \(n\), \(p\), the independence assumption, and the event. For the Poisson approximation set \(\lambda=np\).
Use the binomial probability or sum the required binomial terms. Check that the event is within the binomial support \(0,\ldots,n\).
Use the Poisson probability or sum the required Poisson terms, using the same event and \(\lambda=np\).
Compute the specified error measure and compare it with the tolerance. Use the exact result if the approximation fails or if the problem demands an exact probability.
Why the Event Can Change the Comparison
A useful way to see how the two point probabilities differ is to divide the binomial probability by the Poisson probability. For \(0<p<1\), \(\lambda=np>0\), and \(0\leq k\leq n\), cancellation gives
The first factor reflects the finite number of distinct trials: for \(k>0\), the descending product is generally smaller than \(n^k\). The remaining factors involve the probability of failures and the Poisson exponential term. Their balance depends on \(k\), not merely on \(np\). For a cumulative event, the pointwise differences at its included values are added, so their effects may reinforce or offset one another.
The ratio is stated only for \(0<p<1\), where both distributions assign positive probability to every \(k\) in the binomial support. At \(p=0\), \(X=0\) with probability one and \(\lambda=0\); handle this degenerate case directly rather than forming a ratio with a zero denominator. At \(p=1\), \(X=n\) with probability one, while the Poisson approximation has parameter \(n\); compare the event probabilities directly. For \(k>n\), the exact binomial probability is zero, even though a Poisson distribution can assign positive probability there.
Worked Comparisons
Worked Example: A Zero-Alert Probability With a Tight Tolerance
A hypothetical device is checked in 15 independent inspections. Each inspection independently produces an alert with probability \(p=0.04\). Compare the exact and Poisson probabilities of no alerts, and decide whether the Poisson approximation meets an absolute-error limit of \(0.005\). No interest-rate or mortality assumption applies; the model is a binomial count with \(n=15\) and \(p=0.04\).
Method. Calculate \(P(X=0)\) from \(X\sim\operatorname{Binomial}(15,0.04)\), then compare it with \(P(Z=0)\) for \(Z\sim\operatorname{Poisson}(\lambda)\), where
For the exact binomial probability, zero alerts means all 15 inspections produce no alert:
For the Poisson approximation:
The absolute error, computed from the unrounded probabilities, is approximately \(0.0067253\). Since \(0.0067253>0.005\), the Poisson approximation fails the stated tolerance. Use the exact binomial probability for this requirement.
Check. The exact result is smaller than the Poisson result, and their difference exceeds the limit by approximately \(0.0017253\). The conclusion is consistent with the calculated error.
Common mistake avoided: Calling the approximation acceptable just because \(p=0.04\) is small. The event-specific error must still be checked against the stated tolerance.
Worked Example: Exactly Two Events in Many Rare Trials
A hypothetical portfolio has 200 independent exposures, each with probability \(p=0.005\) of generating a specified type of claim during a period. Compare the exact and Poisson probabilities of exactly two claims. The required absolute error is at most \(0.002\). No interest-rate or mortality assumption applies; use a binomial count with \(n=200\) and \(p=0.005\).
Method. The Poisson parameter is
For the exact probability, first evaluate the combination and the success-probability factor:
The remaining factor is \((0.995)^{198}\approx0.370654\). Therefore,
For \(Z\sim\operatorname{Poisson}(1)\),
Using the unrounded probabilities, the absolute error is approximately \(0.0004609\). Since \(0.0004609<0.002\), the approximation meets the required accuracy. Report probabilities and error to four decimal places: \(0.1844\), \(0.1839\), and \(0.0005\), respectively.
Check. The exact and approximate probabilities differ by less than half of one thousandth, which is below the two-thousandths tolerance. The conclusion follows from the error comparison.
Common mistake avoided: Rounding the exact probability too early or miscalculating the binomial coefficient. The coefficient is \(19{,}900\), and the final comparison uses unrounded values.
Worked Example: A Cumulative Event That Fails the Tolerance
A hypothetical screening procedure is applied in 20 independent cases. Each case has probability \(p=0.10\) of a positive result. Compare the exact and Poisson probabilities of at most one positive result, with an absolute-error tolerance of \(0.01\). No interest-rate or mortality assumption applies; use \(X\sim\operatorname{Binomial}(20,0.10)\).
Method. Set \(\lambda=np=20(0.10)=2\). “At most one” includes both zero and one, so calculate both terms under each model.
The exact binomial terms are
Adding them gives
For \(Z\sim\operatorname{Poisson}(2)\), the corresponding terms are
The absolute error is approximately
Because \(0.0142589>0.01\), the approximation does not meet the tolerance. Use the exact binomial cumulative probability. Rounded to four decimal places, the exact and approximate probabilities are \(0.3917\) and \(0.4060\), with error \(0.0143\).
Check. The Poisson cumulative probability is larger, and the difference exceeds the limit. The conclusion is consistent with the calculated values.
Common mistake avoided: Comparing only the probability of exactly one positive result. The requested event includes zero as well, so both terms must be included under both models.
Common Traps and Exam Decisions
Several shortcuts can lead to an unjustified choice. The exact calculation is not automatically required whenever \(n\) is modest, and the Poisson calculation is not automatically reliable whenever \(p\) is small. Decide using the event and the stated accuracy.
- Using \(p\) without checking the event. Small \(p\) is a useful clue, but it does not establish that a particular point probability or cumulative probability meets the required tolerance.
- Using the wrong Poisson parameter. For a binomial model with parameters \(n\) and \(p\), the standard approximation here uses \(\lambda=np\), not \(n\), \(p\), or \(np(1-p)\).
- Changing the event in one model. “At most \(r\)” includes every integer from 0 through \(r\); “at least one” is one minus the probability of zero. Compare the same event under both distributions.
- Rounding before taking a difference. Keep sufficient precision in both probabilities, calculate the error, and round only the reported results.
- Using a ratio at a boundary. The point-probability ratio above requires \(0<p<1\). At \(p=0\) or \(p=1\), calculate probabilities directly.
- Confusing support with approximation. A binomial count cannot exceed \(n\), but a Poisson count can. If the event includes values above \(n\), their exact binomial probability is zero and must not be silently treated as an ordinary binomial term.
The practical conclusion is simple: if a problem specifies a maximum error, calculate or bound the error for the event in question. If it demands an exact answer, use the binomial model. If no tolerance is supplied and exact computation is straightforward, the exact result avoids an unnecessary approximation decision. This event-based judgment is aligned with the SOA VEE Mathematical Statistics syllabus.
Check Your Understanding
For each problem, identify the exact binomial event and compare it with the corresponding Poisson approximation where requested. Assume independent trials with a constant success probability.
- A hypothetical sensor is tested in 30 independent trials, with alert probability \(p=0.02\) per trial. Compare the exact and Poisson probabilities of no alerts and report their absolute difference to four decimal places. (Answer: Exact \(0.98^{30}\approx0.5455\); Poisson \(e^{-0.6}\approx0.5488\); absolute error \(\approx0.0033\).)
- A hypothetical process has 80 independent opportunities for a defect, with probability \(0.01\) per opportunity. Write the exact and Poisson expressions for exactly one defect, using \(\lambda=np\), and state which expression is the approximation. (Answer: Exact \(\binom{80}{1}(0.01)(0.99)^{79}=0.8(0.99)^{79}\); approximation \(e^{-0.8}(0.8)\).)
- In 50 independent hypothetical cases, each has probability \(0.04\) of a specified event. Compare the exact and Poisson probabilities of at least one occurrence; give the absolute error to four decimal places. (Answer: Exact \(1-(0.96)^{50}\approx0.8701\); Poisson \(1-e^{-2}\approx0.8647\); absolute error \(\approx0.0054\).)
- A hypothetical count has \(n=500\) independent trials and \(p=0.002\). For exactly two successes, use the Poisson approximation and determine whether its absolute error is below \(0.0002\). (Answer: \(\lambda=1\); Poisson probability \(e^{-1}/2\approx0.1839\); exact probability \(\binom{500}{2}(0.002)^2(0.998)^{498}\approx0.1841\); error is approximately \(0.0002\), more precisely about \(0.0002\) and below \(0.0002\).)
- A hypothetical test is given to 10 independent subjects, each with positive-result probability \(0.40\). For the event of zero positive results, state the exact and Poisson expressions and decide whether the approximation meets an absolute-error limit of \(0.01\). (Answer: Exact \((0.60)^{10}\approx0.0060\); Poisson \(e^{-4}\approx0.0183\); absolute error \(\approx0.0123\), so it does not meet the limit.)