From a Binomial Count to a Range Probability
A question about a binomial count may ask for a range, such as the probability of between two and five claims, or a lower tail, such as the probability of no more than four events. Once a binomial count is approximated by a Poisson count, the requested probability is found by adding the Poisson probabilities for the integer outcomes in the event. Getting the endpoints right is as important as calculating the individual terms.
Suppose \(X\) counts successes in \(n\) independent trials, each with the same success probability \(p\). Then \(X\sim\operatorname{Binomial}(n,p)\). When the rare-event approximation is appropriate, use a Poisson random variable \(N\) with parameter \(\lambda=np\). The approximation is useful when \(n\) is relatively large and \(p\) is relatively small; it is not an exact replacement for the binomial model.
The Poisson probabilities are added because the individual count outcomes are mutually exclusive. For example, the event \(2\leq N\leq5\) consists of the four disjoint outcomes \(N=2,3,4,5\). Its probability is the sum of those four point probabilities, not the probability at either endpoint alone.
Translating Event Wording into Integer Endpoints
Count variables take integer values, so ordinary language must be translated carefully. “At most \(b\)” means \(0\leq N\leq b\). “Fewer than \(b\)” means \(0\leq N\leq b-1\). “From \(a\) through \(b\), inclusive” means \(a\leq N\leq b\). These translations determine exactly which terms belong in the sum.
| Wording | Integer event | Poisson sum |
|---|---|---|
| At most \(b\) | \(0\leq N\leq b\) | \(\sum_{k=0}^{b}P(N=k)\) |
| Fewer than \(b\) | \(0\leq N\leq b-1\) | \(\sum_{k=0}^{b-1}P(N=k)\) |
| From \(a\) through \(b\), inclusive | \(a\leq N\leq b\) | \(\sum_{k=a}^{b}P(N=k)\) |
For a lower tail, add from zero through the upper endpoint. For a bounded range, include both endpoints when the wording says “inclusive,” “through,” or “from ... to ...” and the context indicates both are included. If the event is described with strict inequalities, translate those inequalities before calculating. For example, an integer count satisfying \(2<N<6\) must be 3, 4, or 5.
Why Summing the Poisson Terms Works
The exact probability of a binomial count falling in an inclusive integer range is itself a sum of point probabilities. The approximation replaces each binomial point probability by its corresponding Poisson point probability, while keeping the same integer outcomes.
The first equality follows by adding probabilities of disjoint events \(X=k\). The approximation applies the binomial-to-Poisson approximation to each point probability in the finite range. The lower-tail version follows by using outcomes \(0,1,\ldots,b\):
This reasoning does not make the Poisson result exact: the binomial model has parameters \(n,p\), while the approximating Poisson model has parameter \(\lambda=np\). The answer is an approximation to the original binomial event, not a claim that the underlying count has changed distribution.
Efficient Calculation of a Range
A direct calculation uses the Poisson probability formula for each integer. Alternatively, successive terms can be calculated by the recurrence \(P(N=k+1)=P(N=k)\lambda/(k+1)\). This follows by dividing the formula for \(P(N=k+1)\) by the formula for \(P(N=k)\). Once one term is known, the recurrence gives the next term without evaluating a new factorial.
For a binomial model with \(n\) trials and success probability \(p\), calculate \(\lambda=np\).
Write the full list or range of integer counts included. Check whether each endpoint is included.
Use the Poisson probability formula, or use the recurrence to obtain successive terms.
Keep enough precision during the calculation, then round the final probability to the requested number of decimals.
Worked Example: A Bounded Range of Equipment Alerts
A hypothetical monitoring system checks 200 independent units during a shift. Each unit has probability \(0.02\) of producing an alert. Approximate the probability that the shift produces from 2 through 6 alerts, inclusive. No interest rate or mortality assumption is involved; the count model is binomial with \(n=200\) and \(p=0.02\).
Method and parameter. Let \(X\) be the number of alerts. Under the stated independent, equal-probability model, \(X\sim\operatorname{Binomial}(200,0.02)\). The Poisson approximation has parameter \[ \lambda=np=200(0.02)=4. \] Thus use \(N\sim\operatorname{Poisson}(4)\).
Translate the event. “From 2 through 6, inclusive” means the five outcomes \(N=2,3,4,5,6\). The required approximation is their sum.
Calculate and add. Using \(P(N=k)=e^{-4}4^k/k!\), the terms are \(P(N=2)=0.1465251\), \(P(N=3)=0.1953668\), \(P(N=4)=0.1953668\), \(P(N=5)=0.1562935\), and \(P(N=6)=0.1041956\). Therefore,
The approximation is \(0.7977\) to four decimal places. As a check, the terms are all nonnegative, and their sum is less than 1. The common mistake avoided here is including only the endpoints or omitting one of the interior counts.
Worked Example: A Lower Tail of Service Requests
A hypothetical service desk receives 500 independent requests during a defined period, with probability \(0.006\) that any one request requires escalation. Approximate the probability that at most three requests require escalation. No interest rate or mortality assumption is involved; \(X\) is binomial with \(n=500\) and \(p=0.006\).
Method and parameter. The Poisson parameter is \[ \lambda=np=500(0.006)=3, \] so use \(N\sim\operatorname{Poisson}(3)\). “At most three” includes 0, 1, 2, and 3; it is not the event \(N<3\).
The separate Poisson terms are \(P(N=0)=0.04978707\), \(P(N=1)=0.14936121\), \(P(N=2)=0.22404181\), and \(P(N=3)=0.22404181\). Their unrounded sum is \(13e^{-3}=0.64723190\). The approximate probability is \(0.6472\) to four decimal places. The check is that the event includes exactly four possible nonnegative counts, from 0 to 3.
The method is a lower-tail sum. Its advantage is that it uses only the terms in the requested event; there is no need to calculate probabilities for larger counts and subtract them from 1.
Worked Example: A Bounded Range with a Larger Mean
A hypothetical inspection process examines 300 independent components. Each component has probability \(0.02\) of needing a second inspection. Approximate the probability that between 5 and 8 components, inclusive, need a second inspection. No interest rate or mortality assumption is involved; \(X\sim\operatorname{Binomial}(300,0.02)\).
Method and parameter. Here \[ \lambda=np=300(0.02)=6, \] so the approximating distribution is \(N\sim\operatorname{Poisson}(6)\). The inclusive range contains \(5,6,7,8\), hence four terms.
Start with \(P(N=5)=e^{-6}6^5/5!=0.1606231\). Apply the recurrence to calculate each following term:
Adding the four probabilities gives \[ 0.1606231+0.1606231+0.1376770+0.1032578=0.5621810. \] Therefore, the Poisson approximation is \(0.5622\) to four decimal places. The result is positive and less than 1, as a probability must be. The common mistake avoided is using a recurrence factor of \(\lambda/k\) with the old index \(k\); when moving from \(k\) to \(k+1\), the denominator is \(k+1\).
Common Endpoint Trap
Cumulative probabilities are useful for ranges, but the subtraction must match the endpoints. If \(F_N(b)=P(N\leq b)\), then for integers \(a\leq b\),
The subtraction removes all outcomes below \(a\), not outcomes through \(a\). For example, for a Poisson variable with \(\lambda=3\), the event \(2\leq N\leq5\) has probability \(0.7169338\), obtained by adding the terms for 2, 3, 4, and 5. A wrong calculation is \(F_N(5)-F_N(2)\), which removes the outcomes 0, 1, and 2 and leaves only counts 3, 4, and 5. That gives \(0.4928920\), so it omits the required \(N=2\) term. The correction is \(F_N(5)-F_N(1)=0.9160821-0.1991483=0.7169338\).
Check Your Understanding
For each problem, use a Poisson approximation and give the requested probability to four decimal places.
- A hypothetical sensor network has 250 independent sensors, each with probability \(0.012\) of reporting a fault. Approximate the probability of from 1 through 4 faults, inclusive. (Answer: \(0.7655\))
- A hypothetical dispatch center handles 400 independent calls, each having probability \(0.01\) of requiring a specialist. Approximate the probability of at most 2 specialist calls. (Answer: \(0.2381\))
- A hypothetical packaging line produces 600 independent packages, each with probability \(0.005\) of a seal defect. Approximate the probability of from 2 through 5 defects, inclusive. (Answer: \(0.7169\))
- A hypothetical laboratory tests 1,000 independent samples, each with probability \(0.004\) of requiring repeat analysis. Approximate the probability that at least 3 but fewer than 7 samples require repeat analysis. (Answer: \(0.6512\))