Approximating a Single Binomial Probability

Why Approximate a Single Binomial Probability?

A binomial model is natural when a fixed number of independent trials each produce either a success or a failure with the same success probability. Its exact point probability can require a large binomial coefficient and powers of probabilities. When there are many trials and success is relatively unlikely, a Poisson probability can provide a simpler approximation for the chance of exactly $k$ successes.

The approximation does not replace the model assumptions or change the event. It estimates the same point probability, $P(X=k)$, using a Poisson random variable whose mean matches the binomial mean. A sound solution states both the binomial parameters and the Poisson parameter, computes the approximation, and compares it with the exact binomial value when requested.

Definition: If $X\sim\operatorname{Binomial}(n,p)$, then $X$ counts successes in $n$ independent trials, each with success probability $p$. For an integer $k$ with $0\leq k\leq n$, the exact point probability is $$ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}. $$ When a Poisson approximation is appropriate, use $Z\sim\operatorname{Poisson}(\lambda)$ with $\lambda=np$. The approximation to the binomial point probability is $$ P(X=k)\ \approx\ P(Z=k)=e^{-np}\frac{(np)^k}{k!}. $$

The approximation is most useful when $n$ is sufficiently large, $p$ is small, and the product $np$ is moderate. These are practical conditions, not a guarantee that every individual probability will be close. The exact binomial formula remains the reference value when an exact comparison is required. If $k>n$, the binomial probability is zero, so a positive Poisson point probability cannot be a useful approximation to it.

How to Approximate One Point Probability

Keep the event fixed throughout the calculation: if the question asks for exactly $k$ successes, use that same $k$ in both models. First calculate $np$, then substitute this value for $\lambda$ in the Poisson mass function. If the question asks for a comparison, calculate the binomial probability separately; do not call the Poisson value the exact answer.

State the binomial model.
Identify the number of trials $n$, the success probability $p$, and the requested count $k$. Confirm that the trials are independent and have a common success probability.

Match the mean.
Calculate $\lambda=np$. This is the mean of the approximating Poisson distribution.

Calculate the Poisson point probability.
Evaluate $e^{-\lambda}\lambda^k/k!$ using the specified $k$.

Calculate the exact value if comparison is needed.
Evaluate $\binom{n}{k}p^k(1-p)^{n-k}$, then compare the two displayed probabilities at the requested precision.

Worked Example: Exactly Three Delayed Deliveries

In a hypothetical shipment of 200 independent parcels, each parcel has probability $0.01$ of being delayed. Approximate the probability that exactly 3 parcels are delayed with a Poisson model, and compare it with the exact binomial probability. Assume the delay indicators are independent and have the same probability for all parcels.

Method. This is a binomial count with $n=200$, $p=0.01$, and $k=3$. The mean is $$ \lambda=np=200(0.01)=2. $$

The Poisson approximation is $$ P(Z=3)=e^{-2}\frac{2^3}{3!} =e^{-2}\frac{8}{6} =0.180447\ldots $$ Thus the Poisson approximation, rounded to four decimal places, is $0.1804$.

For the exact binomial value, $$ P(X=3)=\binom{200}{3}(0.01)^3(0.99)^{197}. $$ The binomial coefficient is $$ \binom{200}{3}=\frac{200(199)(198)}{3(2)(1)}=1{,}313{,}400. $$ Then $$ P(X=3)=1{,}313{,}400(0.000001)(0.99)^{197} =1.3134(0.99)^{197} \approx 0.1814. $$ The exact value is approximately $0.181355$, so both values round to probabilities near $0.18$, with the Poisson estimate slightly lower. The calculation checks the requested count in both formulas: each evaluates the probability of exactly 3 delays.

Common mistake avoided: Using $k=2$ because the mean is 2 would answer a different question. The mean determines $\lambda$; it does not replace the specified count $k=3$.

Why the Substitution Works

The approximation comes from the rare-event limit for a binomial distribution. For a fixed nonnegative integer $k$, let $p=\lambda/n$, so that $np=\lambda$ stays fixed as $n$ grows. Starting from the exact binomial probability and factoring it gives

\begin{aligned} P(X=k) &=\binom{n}{k}\left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k}\\ &=\frac{\lambda^k}{k!} \left(\frac{n(n-1)\cdots(n-k+1)}{n^k}\right) \left(1-\frac{\lambda}{n}\right)^n \left(1-\frac{\lambda}{n}\right)^{-k}. \end{aligned}

For fixed $k$, the first parenthesized factor is the product $(1)(1-1/n)\cdots(1-(k-1)/n)$, which tends to 1 as $n$ grows. The standard exponential limit gives $(1-\lambda/n)^n\to e^{-\lambda}$, while $(1-\lambda/n)^{-k}\to 1$. Therefore the point probability tends to

e^{-\lambda}\frac{\lambda^k}{k!},

the Poisson probability mass function. This derivation explains the formula and its main setting: many trials, small success probability, and a fixed mean. For a finite $n$, it remains an approximation rather than an identity.

Worked Example: Exactly Two Claims in a Large Portfolio

In a hypothetical portfolio, 400 independent policies each have probability $0.01$ of producing a claim during a specified period. Find the Poisson approximation to the probability of exactly 2 claims and compare it with the exact binomial value. Assume the claim indicators are independent and identically distributed.

Method. Here $n=400$, $p=0.01$, and $k=2$. The matched Poisson mean is $$ \lambda=np=400(0.01)=4. $$ The approximate probability is $$ P(Z=2)=e^{-4}\frac{4^2}{2!} =8e^{-4} =0.146525\ldots, $$ or $0.1465$ to four decimal places.

The exact binomial probability is $$ P(X=2)=\binom{400}{2}(0.01)^2(0.99)^{398}. $$ Calculate the coefficient and success-probability factor: $$ \binom{400}{2}=\frac{400(399)}{2}=79{,}800, \qquad 79{,}800(0.01)^2=79{,}800(0.0001)=7.98. $$ Thus $$ P(X=2)=7.98(0.99)^{398}\approx 0.146154, $$ which rounds to $0.1462$. The exact and approximate values are close, but they are not identical.

Check. The exact calculation uses the probability of two claims and 398 nonclaims; its exponents add to 400. The Poisson calculation uses the same target count $k=2$ and the mean $4$.

Common mistake avoided: The value $np=4$ is not the number of claims requested. Substituting $k=4$ would calculate the probability of exactly four claims rather than exactly two.

Comparison Can Reveal a Poor Approximation

A Poisson calculation is simple, but simplicity alone does not establish accuracy. If $p$ is not small or $n$ is not large, the binomial and Poisson point probabilities can differ noticeably. The comparison is particularly important when the problem asks for both values or when the model conditions are only borderline. State the approximation as an estimate and retain the exact binomial formula as the benchmark.

Worked Example: Zero Equipment Failures

A hypothetical workshop operates 20 independent machines during a shift. Each machine has probability $0.10$ of failing during that shift. Estimate the probability of zero failures with a Poisson model and compare it with the exact binomial probability. Assume each machine has the same failure probability and machine failures are independent.

Method. The binomial parameters are $n=20$, $p=0.10$, and the requested count is $k=0$. The Poisson mean is $$ \lambda=np=20(0.10)=2. $$ The Poisson estimate is $$ P(Z=0)=e^{-2}\frac{2^0}{0!}=e^{-2}=0.135335\ldots, $$ which is $0.1353$ to four decimal places.

For the exact probability, zero failures means all 20 machines do not fail. Hence $$ P(X=0)=\binom{20}{0}(0.10)^0(0.90)^{20} =(0.90)^{20} =0.1215766546\ldots, $$ or $0.1216$ to four decimal places. The two results differ by more than in the earlier examples. Here $p=0.10$ and $n=20$ do not provide as strong a rare-event setting, so the Poisson value should be presented as an approximation, not as the exact probability.

Check. Since $k=0$, the binomial term contains no factor of $p$; it is the probability that all trials are failures. This agrees with the direct expression $(1-p)^n$.

Common mistake avoided: For exactly zero successes, do not calculate $1-P(X=0)$. That complement is the probability of at least one success, a different event.

Exam Traps and Interpretation

The central calculation is short, so errors often come from setting up the model rather than evaluating the exponential. Before substituting values, distinguish the trial count $n$, the per-trial probability $p$, the mean $\lambda=np$, and the requested count $k$. These quantities have separate roles in the formula.

Common exam trap: For $n=400$, $p=0.01$, and $k=2$, using $\lambda=p=0.01$ would give $$ e^{-0.01}\frac{(0.01)^2}{2!}\approx 0.0000495, $$ which is not the Poisson approximation for this binomial model. The correct mean is $\lambda=np=4$, giving $$ e^{-4}\frac{4^2}{2!}\approx 0.1465. $$ Always multiply the number of trials by the success probability before evaluating the Poisson mass.

Also check whether the requested event is a point event. “Exactly $k$” uses one mass $e^{-\lambda}\lambda^k/k!$. An event covering several possible counts requires combining multiple point probabilities and is a different calculation. Keep the solution here focused on one specified value of $k$.

Do not confuse the mean with the observed count. The mean parameter is $np$; $k$ comes from the event in the question.
Do not omit the factorial. The Poisson point mass includes $k!$, including $0!=1$.
Do not label an estimate as exact. The binomial formula gives the exact value under the stated model; the Poisson formula estimates it.
Check the support. A binomial count cannot exceed $n$, even though a Poisson model assigns positive probability to every nonnegative integer.
Keep the model assumptions visible. If trials are dependent or have different success probabilities, the stated binomial model may not apply.

For actuarial frequency work, the same calculation can approximate a count of rare events across many independent exposure units, such as hypothetical policy-level claim indicators. The interpretation depends on the unit of exposure and period specified in the model: $p$ is the probability for one trial over that period, and $np$ is the expected count over all $n$ trials. Mixing periods or using a probability that does not match the unit changes the approximation.

Key takeaway: For one binomial point probability, first identify $n$, $p$, and $k$, then set $\lambda=np$ and evaluate $e^{-\lambda}\lambda^k/k!$. If asked to compare, calculate $\binom{n}{k}p^k(1-p)^{n-k}$ separately and report the Poisson result as an approximation.

Check Your Understanding

For each problem, use the Poisson model with $\lambda=np$. Round probabilities to four decimal places. Where requested, compare with the exact binomial probability.

A hypothetical inspection covers 300 independent components, each with probability $0.005$ of a defect. Approximate the probability of exactly 2 defects. (Answer: $\lambda=1.5$; Poisson approximation $0.2510$.)
In a hypothetical group of 120 independent applications, each has probability $0.01$ of requiring additional review. For exactly zero reviews, give both the exact binomial probability and the Poisson approximation. (Answer: exact $(0.99)^{120}\approx0.2994$; approximation $e^{-1.2}\approx0.3012$.)
A hypothetical set of 250 independent devices has probability $0.008$ per device of a specified fault. Approximate the probability of exactly 3 faults and compare with the exact binomial value. (Answer: $\lambda=2$; approximation $0.1804$; exact $\binom{250}{3}(0.008)^3(0.992)^{247}\approx0.1812$.)
A hypothetical fleet has 40 independent vehicles, each with probability $0.05$ of needing a particular repair during the month. Approximate the probability that exactly 5 vehicles need that repair. (Answer: $\lambda=2$; approximation $e^{-2}2^5/5!\approx0.0361$.)
For 30 independent hypothetical accounts, the probability of a specified event for any one account is $0.08$. Approximate the probability that exactly 4 accounts experience the event. (Answer: $\lambda=2.4$; approximation $e^{-2.4}(2.4)^4/4!\approx0.1254$.)

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