Computing Exact Poisson Point Probabilities

Introduction

A question about an exact count is narrower than a question about a range of counts. If a hypothetical service desk averages a known number of incoming requests per hour, the probability of receiving exactly a specified number during a shift is a single Poisson point probability. The calculation is short, but it is easy to use the wrong rate, mishandle the factorial, or answer a different question from the one asked.

This tutorial focuses on calculating that one probability from the Poisson probability mass function. The count model and the interpretation of its rate are assumed from the earlier material on Poisson count models and exposure. The calculations here are aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: If $X$ is a Poisson random variable with parameter $\lambda>0$, then, for an integer $k\geq 0$, $$ P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}. $$ Here, $\lambda$ is the expected count over the specified observation period, $k$ is the exact count of interest, and $k!=k(k-1)\cdots 2\cdot 1$, with $0!=1$.

Set the Parameter Before Calculating

The Poisson parameter $\lambda$ must describe the whole observation period in the question. If events occur at a constant rate $r$ per exposure unit, and the observed exposure is $E$ in those same units, then

\lambda=rE.

For example, a rate stated per machine-day must be multiplied by the number of machine-days observed, not merely by the number of calendar days. If the observation period is given in a different time unit from the rate, convert one of them before multiplying. Once $\lambda$ is found, use it—not the rate per unit by itself—in the probability mass function.

The model assumes that the count $X$ is Poisson over the specified exposure, with the stated parameter. In applications, this is appropriate when the count is modeled as arising at a constant rate over the period and the Poisson count assumptions are reasonable. If the rate changes materially over the period, or the events do not fit the assumed count model, the formula may not apply without further modeling.

Match the units.
Express the event rate and the observed exposure in compatible units.

Find the period parameter.
Multiply the rate per exposure unit by the total exposure to obtain $\lambda$.

Substitute the exact count.
Use the requested integer $k$ in $e^{-\lambda}\lambda^k/k!$.

Evaluate and round.
Keep sufficient precision during the calculation and round the final probability as requested.

Why the Point-Probability Formula Has This Form

A brief justification helps explain the separate factors in the formula. In a Poisson process with rate $r$, let $p_k(t)$ be the probability of exactly $k$ events by time $t$. Over a short additional interval, the probability of one event is approximately $r\,\Delta t$, the probability of no event is approximately $1-r\,\Delta t$, and the probability of two or more events is negligible compared with $\Delta t$. Consequently, for $k\geq 1$,

p_k'(t)=r p_{k-1}(t)-r p_k(t).

The first term accounts for having $k-1$ events and then one more; the second accounts for having $k$ events and then none. For $k=0$, $p_0'(t)=-rp_0(t)$, with $p_0(0)=1$, so $p_0(t)=e^{-rt}$. Substitution into the recurrence, successively for $k=1,2,\ldots$, gives

p_k(t)=e^{-rt}\frac{(rt)^k}{k!}.

The factor $rt$ is the expected count over the period, so writing $\lambda=rt$ gives the stated Poisson probability mass function. The exponential term reflects the probability of no events over the period, while the power and factorial determine the probability for the specified count.

Worked Point Probabilities

Worked Example: Exact Request Count During a Shift

A hypothetical support desk receives requests at a constant modeled rate of $2.4$ requests per hour. Find the probability of exactly $2$ requests during a $90$-minute shift. No interest or mortality assumption is involved; the count model is $X\sim\operatorname{Poisson}(\lambda)$.

Method and parameter. Convert the shift length to hours: $90/60=1.5$ hours. The expected count during the shift is

\lambda=(2.4\text{ requests/hour})(1.5\text{ hours})=3.6\text{ requests}.

Thus $X\sim\operatorname{Poisson}(3.6)$, and the requested exact count is $k=2$. Substitute into the point-probability formula:

\begin{aligned} P(X=2) &=e^{-3.6}\frac{3.6^2}{2!}\\ &=e^{-3.6}\frac{12.96}{2}\\ &=6.48e^{-3.6}\\ &\approx 6.48(0.02732372245)\\ &\approx 0.1770577215. \end{aligned}

Rounded to four decimal places, the probability is 0.1771. The result is between $0$ and $1$, as a probability must be. The method avoids using the hourly rate $2.4$ as though it were the parameter for the full $1.5$-hour shift.

Worked Example: A Factorial and a Rate Per Machine-Day

A hypothetical inspection system records events at a modeled rate of $0.08$ events per machine-day. It monitors $3$ machines for $5$ days. Find the probability of exactly $4$ events over all the monitored exposure. No interest or mortality assumption is involved; assume the total count is Poisson.

Method and parameter. The exposure is $3(5)=15$ machine-days, so

\lambda=(0.08\text{ events/machine-day})(15\text{ machine-days})=1.2.

Therefore $X\sim\operatorname{Poisson}(1.2)$. For exactly $4$ events, $4!=4(3)(2)(1)=24$, and $1.2^4=2.0736$. Hence

\begin{aligned} P(X=4) &=e^{-1.2}\frac{1.2^4}{4!}\\ &=e^{-1.2}\frac{2.0736}{24}\\ &=0.0864e^{-1.2}\\ &\approx 0.0864(0.3011942119)\\ &\approx 0.0260231799. \end{aligned}

Rounded to four decimal places, the probability is 0.0260. The calculation uses the full $15$ machine-days of exposure and the factorial for the requested count of $4$. The main mistake avoided is multiplying the rate by $5$ calendar days while ignoring that $3$ machines were monitored.

Worked Example: Exact Count With a Larger Factorial

A hypothetical maintenance team models equipment alerts at a constant rate of $0.8$ alerts per operating day. Find the probability of exactly $5$ alerts during $5$ operating days. No interest or mortality assumption is involved; assume the count over the period follows a Poisson distribution.

Method and parameter. The expected number of alerts is

\lambda=(0.8\text{ alerts/day})(5\text{ days})=4.

Thus $X\sim\operatorname{Poisson}(4)$. The requested count is $k=5$, so $5!=120$ and $4^5=1024$. Substitution gives

\begin{aligned} P(X=5) &=e^{-4}\frac{4^5}{5!}\\ &=e^{-4}\frac{1024}{120}\\ &=e^{-4}\frac{128}{15}\\ &\approx (0.01831563889)(8.533333333)\\ &\approx 0.1562934519. \end{aligned}

Rounded to four decimal places, the probability is 0.1563. As a check, the factorial division gives $1024/120=8.533333333\ldots$, not $1024/5=204.8$. Using $k$ rather than $k!$ in the denominator would greatly overstate the point probability.

Common Mistakes and Checks

A point probability is the probability of one particular count. Keep the requested event visible throughout the work: $P(X=4)$, for example, is not the probability of at most four or at least four events. Calculating a range or tail requires additional steps and is a different question.

A common error is to use the rate per unit of exposure where the total-period parameter belongs. Suppose a hypothetical count has rate $0.5$ per day over $4$ days and the question asks for exactly $2$ events. The incorrect calculation uses $\lambda=0.5$:

e^{-0.5}\frac{0.5^2}{2!}\approx 0.0758.

The correction is to use $\lambda=(0.5)(4)=2$, giving

P(X=2)=e^{-2}\frac{2^2}{2!} =2e^{-2} \approx 0.2707.

The wrong result answers a different model: a one-day period, not the stated four-day period. Other useful checks are to confirm $k$ is a nonnegative integer, evaluate $k!$ rather than $k$, and postpone rounding until the final probability. For $k=0$, the convention $0!=1$ gives $P(X=0)=e^{-\lambda}$; no separate factorial adjustment is needed.

Key takeaway: First compute the expected count $\lambda$ for the full exposure. Then evaluate $e^{-\lambda}\lambda^k/k!$ for the exact integer $k$, checking units and factorial arithmetic before rounding.

Check Your Understanding

For each hypothetical scenario, identify the period parameter and calculate the probability of the stated exact count. Round probabilities to four decimal places.

A monitoring service models $1.5$ alerts per day. Find the probability of exactly $0$ alerts in $2$ days. (Answer: $0.0498$)
A test station models $0.4$ faults per operating hour. Find the probability of exactly $3$ faults during $5$ hours. (Answer: $0.1804$)
A hypothetical device fleet has a modeled rate of $6$ events per $1{,}000$ device-months. For $500$ devices monitored for $2$ months, find the probability of exactly $2$ events. (Answer: $0.0446$)
A hypothetical group of $8$ machines is monitored for $3$ days. The modeled rate is $0.25$ events per machine-day. Find the probability of exactly $3$ events. (Answer: $0.0892$)
A hypothetical monitoring system models $1.4$ alarms per day over $5$ days. Find the probability of exactly $6$ alarms. (Answer: $0.1490$)

What You'll Learn