The Poisson Probability Distribution
ADA University, School of Business
Information Communication Technologies Agency, Statistics Unit
2025-11-05
Ruslan Muslumov, PhD
ADA University
🤔 Think (1-2 minutes):
Consider events that occur randomly over time or space. Can you think of three examples from your daily life where events happen with some average rate but at unpredictable times?
👥 Pair (2-3 minutes):
Share your examples with a partner. Discuss:
🗣️ Share:
Let’s hear a few examples from the class and discuss how they relate to today’s topic.
Definition
The Poisson probability distribution is a discrete probability distribution that describes the number of events that occur within a fixed interval of time or space, given a known average rate of occurrence. It is named after the French mathematician Siméon Denis Poisson.
Description
Split up the time period into (n) subintervals so that:
Remark
Then the total number of events in the given time interval is just the total number of subintervals that contain one event.
Letting (= np) and taking the limit of the binomial probability
[ p(y) = p^y(1 - p)^{n-y} ]
as (n ), we have
[ \[\begin{align} \lim_{n \to \infty} \binom{n}{y} p^y(1 - p)^{n-y} &= \lim_{n \to \infty} \frac{n(n - 1) \cdots (n - y + 1)}{y!} \left( \frac{\lambda}{n} \right)^y \left(1 - \frac{\lambda}{n}\right)^{n-y} \end{align}\] ]
[ \[\begin{align} &= \lim_{n \to \infty} \frac{\lambda^y}{y!} \left(1 - \frac{\lambda}{n}\right)^{n} \frac{n(n - 1) \cdots (n - y + 1)}{n^y} \left(1 - \frac{\lambda}{n}\right)^{-y} \end{align}\] ]
[ \[\begin{align} &= \frac{\lambda^y}{y!} \lim_{n \to \infty} \left(1 - \frac{\lambda}{n}\right)^{n} \cdot \lim_{n \to \infty} \left(1 - \frac{\lambda}{n}\right)^{-y} \cdot \lim_{n\to \infty}\left((1-\frac{1}{n}) \cdots (1 - \frac{y-1}{n})\right) \end{align}\] ]
[ = ]
Definition
A random variable (Y) is said to have a Poisson distribution if and only if
[ p(y) = e^{-}, y = 0,1,2,,> 0 ]
Example
Show that the probabilities assigned by the Poisson probability distribution satisfy the requirements that (0 p(y) ) for all (y) and
[ _{y}p(y) = 1 ]
Proof
Because (> 0), it is obvious that (p(y) > 0) for (y = 0, 1, 2, ), and that (p(y) = 0) otherwise.
Further,
[ {y=0}^{} p(y) = {y=0}^{} e^{-} = e^{-} _{y=0}^{} = e^{-} e^{} = 1 ]
because the infinite sum (_{y=0}^{} ) is a series expansion of (e^).
Question
Before we move to applications, can you identify:
Answers
Example
Suppose that a random system of police patrol is devised so that a patrol officer may visit a given beat location (Y = 0, 1, 2, 3, ) times per half-hour period, with each location being visited an average of once per time period. Assume that (Y) possesses, approximately, a Poisson probability distribution.
Calculate the probability that the patrol officer will:
Solution
For this example, the time period is a half-hour, and the mean number of visits per half-hour interval is (= 1). Then
[ p(y) = , y = 0, 1, 2, ]
The event that a given location is missed in a half-hour period corresponds to (Y = 0), and
[ P(Y = 0) = p(0) = = e^{-1} ]
Solution (continued)
Similarly,
[ p(1) = = e^{-1} ]
and
[ p(2) = = ]
The probability that the location is visited at least once is the event (P(Y )). Then
[ P(Y ) = _{y=1}^{} p(y) = 1 - p(0) = 1 - e^{-1} ]
Example
A certain type of tree has seedlings randomly dispersed in a large area, with the mean density of seedlings being approximately five per square yard. If a forester randomly locates ten 1-square-yard sampling regions in the area, find the probability that none of the regions will contain seedlings.
Solution
(Y), the number of seedlings per square yard, can be modeled as a Poisson random variable with (= 5). Thus,
[ P(Y = 0) = p(0) = = e^{-5} ]
The probability that (Y = 0) on ten independently selected regions is
[ (e{-5}){10} ]
because the probability of the intersection of independent events is equal to the product of the respective probabilities. The resulting probability is extremely small.
Applications in Nature and Society
The Poisson distribution appears in many contexts:
Approximation Conditions
The Poisson distribution can be used to approximate the binomial distribution when:
Why It Works
When these conditions hold, the binomial distribution with parameters (n) and (p) is approximately Poisson with parameter (= np).
Example
Suppose that (Y) possesses a binomial distribution with (n = 20) and (p = 0.1). Find the exact value of (P(Y )) using the table of binomial probabilities, Table: “Binomial Distribution Histograms”. Use the following table, to approximate this probability, using a corresponding probability given by the Poisson distribution. Compare the exact and approximate values for (P(Y )).
Reference Tool
Solution
Exact Binomial Calculation:
According to Table, the exact value of (P(Y ) = 0.867).
Poisson Approximation:
If (W) is a Poisson-distributed random variable with (= np = 20 = 2), previous discussions indicate that (P(Y )) is approximately equal to (P(W )). Table 3, Appendix 3, gives
[ P(W ) = 0.857 ]
Comparison: The Poisson approximation is quite good, yielding a value that differs from the exact value by only 0.01 (about 1% relative error).
Poisson Probabilities Table
Theorem
If (Y) is a random variable possessing a Poisson distribution with parameter (), then
[ = E(Y) = ^2 = V(Y) = ]
Key Insight
In the Poisson distribution, the mean equals the variance. This is a unique property that distinguishes it from other distributions!
Example
Industrial accidents occur according to a Poisson process with an average of three accidents per month. During the last two months, ten accidents occurred.
Questions:
Solution
The number of accidents in two months, (Y), has a Poisson probability distribution with mean (= 2 = 6). The probability that (Y) is as large as 10 is
[ P(Y ) = _{y=10}^{} ]
By Poisson Distribution Applet or from the theorem:
[ = = 6, ^2 = = 6, = = 2.45 ]
Solution (continued)
The empirical rule tells us that we should expect (Y) to take values in the interval () with a high probability.
Notice that:
[ + 2= 6 + (2)(2.45) = 10.90 ]
Interpretation: The observed number of accidents, (Y = 10), does not lie more than (2) from (), but it is close to the boundary. Thus, the observed result is not highly improbable, but it may be sufficiently improbable to warrant an investigation.
🤔 Think (2 minutes):
You are the safety manager. Based on the analysis showing 10 accidents in 2 months (close to the (+ 2) boundary), would you:
👥 Pair (3 minutes):
Discuss with your partner:
📝 Summary
Theoretical Foundation:
Practical Properties:
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📧 Questions?
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Course Resources:

Mathematical Statistics - The Poisson Probability Distribution