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1641 LearnersLast updated on November 24, 2025

Poisson distribution is a discrete probability distribution. It gives the probability of an event that might occur a particular number of times in a given time period. Poisson distribution can be used in many real life scenarios. For example, to find the number of emails you might receive in an hour. Let’s learn more about this in this topic.
A Poisson distribution is used to predict or estimate the number of times an event might occur within a given period of time. This type of distribution method is specifically used when the variables are discrete count variables. We usually use Poisson distribution when dealing with variables, such as economic and financial data.
The formula used to calculate the probability of an event occurring discreetly over a given period of time is:
\(P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}\)
Where:
e is approximately 2.718 (Euler’s number),
λ is the average number of events in the interval
k! is the factorial of k
k is the actual number of occurrences.
In a Poisson distribution, both the mean and variance are equal to λ. Here, λ is greater than 0.
Imagine you are waiting for text messages from your friends. Based on your history, you know that on average, you receive three text messages per hour.
Even though the average is 3, you know that in any specific hour, the actual number could vary. You might get 0, 3, or 5 texts. The text messages arrive independently of one another (receiving one doesn't change the chance of receiving another).
This is a classic Poisson Distribution problem because:
How the Probability Looks:
Using the Poisson distribution, we can calculate the likelihood of receiving a specific number of texts in the next hour.
Calculation:
\(P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}\)
Where:
Example Calculation for three texts:
\(P(X=3) = \frac{3^3 \cdot e^{-3}}{3!} = \frac{27 \cdot 0.0498}{6} \approx 0.22\)
Use the Poisson distribution calculator to check.
For a scenario to be modeled as a Poisson distribution, it must satisfy these specific properties (often called the “Assumptions of the Poisson Process”):
Here are the key characteristics of the Poisson Distribution:


In a Poisson distribution, the average and the spread are identical. They are both defined by the single parameter \lambda (Lambda).
Example:
If your average rate of texts is \(\lambda = 4:\)
A Poisson Distribution Table is a reference chart used to find probabilities without doing complex calculations manually. Think of it as a “Cheat Sheet” for the Poisson formula.
Instead of typing \(P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}\) into a calculator every time, you simply look up the value where a specific Row and Column meet.
There are two distinct types of tables. It is critical to know which one you are using, as they give very different answers.
This table gives the probability of an event occurring an exact number of times.
This table gives the probability of an event occurring up to and including a certain number of times. It is the running total of the probabilities.
A Poisson distribution graph visualizes the probability of a certain number of events occurring within a fixed interval (like time or space).
Here are the key features of this graph:
Discrete Data (Bars or Dots): Unlike a Bell Curve (Normal distribution) which is a smooth line, a Poisson graph is "discrete." This means it uses bars or dots because you count whole events (e.g., 0 calls, 1 call, 2 calls); you cannot have 2.5 calls.
The Axes:
The Shape (Skewness):
|
Feature |
Poisson Distribution |
Normal (Gaussian) Distribution |
| Data Type | Discrete (Whole numbers only) | Continuous (Any real number) |
| Parameters | One: \(\lambda\) (Lambda). This represents both the Mean and Variance. | Two: \(\mu\) (Mean) and \(\sigma\) (Standard Deviation). |
| Shape | Skewed. It leans to the right (positive skew) when \(\lambda\) is small. | Symmetric. Always a perfect bell curve shape. |
| Bounds | Bounded. Cannot go below 0 (you can't have -1 phone calls). | Unbounded. Goes from \(-\infty\) to \(+\infty.\) |
| Peak | Peak changes position as \(\lambda\) increases. | Peak is always exactly at the mean (\(\mu\)). |
Poisson distribution is important because of its wide application. Here are few other reasons why it is such a prominent tool in mathematics:
It can be quite confusing when learning about Poisson distribution. Here are a few tips and tricks students can use to master Poisson distribution.
When solving problems involving Poisson Distribution, students can make quite a few mistakes, which may lead to incorrect answers. So here are a few common mistakes and how to avoid them:
Poisson distribution is widely used in the real world. Here are a few examples where we use Poisson Distribution.
An email server receives an average of λ = 2 emails per minute. What is the probability of receiving exactly 3 emails per minute?
The probability of receiving exactly 3 emails per minute ≈ 0.1804.
\(P(x) = (X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
P(3) = (X = k) ≈ 0.1804
Identify λ = 2 and k = 3, also remember that e is Euler's constant (2.718)
Substitute the values into the formula
Calculate\( 2^3 = 8\) and 3! = 6.
Use \(e^{-2} ≈ 0.1353\) and calculate the final probability.
A call center receives an average of 4 calls per hour. What is the probability of receiving exactly 5 calls in a 2 hour period?
For 2 hours ≈ 0.0917.
For 2 hours, λ = 4 × 2 = 8.
P(5) = (X = k) ≈ 0.0917
First, convert the rate of calls: 4 calls per hour × 2 hours = 8 calls in 2 hours.
Substitute k = 5
Calculate \(8^5\) = 32768 and 5! = 120.
Use \(e^{-8} ≈ 0.000335\) (where e is 2.718) and calculate the final probability.
In a large production of 1000 items, each item has a 0.005 probability of being defective. Use the Poisson approximation to find the probability of finding exactly 3 defective items.
The probability of finding exactly 3 defective items ≈ 0.1404.
Calculate λ = 1000 × 0.005 = 5.
Then,
\(P(x) = (X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
\( P(3) = (X = k) = \frac{e^{-5} \times 5^3}{3!}≈{{0.00674 \times 125} \over{6}}≈{0.8425 \over6}≈ 0.1404 \)
Two independent call centers receive calls with an average of 3 and 2 per hour respectively. What is the probability that the total number of calls in one hour is exactly 4?
The probability is ≈ 0.1755.
Sum of independent Poisson variables = λ = 3 + 2 = 5
\(P(x) = (X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
\( P(4) = (X = k) =\frac{e^{-5} \times 5^4}{4!}≈{0.00674 \times625 \over24}≈{4.2124 \over24}≈ 0.1755 \)
Add the independent variables 3 + 2 = 5 = λ
Substitute the values λ = 5 and k = 4
Calculate \(e^{-5}\) and find the final probability.
A supermarket receives an average of 3 customer complaints per day. What is the probability of receiving exactly 5 complaints in one day?
The probability of receiving exactly 5 complaints is ≈ 0.1008.
\(P(x) = (X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
\(P(5) = (X = k) = \frac{e^{-3} \times 3^5}{5!}≈{0.0498 \times243 \over120}≈{12.1014 \over120}≈ 0.1008\)
Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!






