1536 LearnersLast updated on 8 September 2025
Poisson Distribution
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.
What is Poisson Distribution?
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) = (e^(-λ) * λ^k) / 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.
Properties of Poisson Distribution
Knowing about the key properties will help us utilize Poisson distribution better. Here are a few important properties:
- Poisson distribution is discrete, which means we will be dealing with non-continuous, countable values. This is because, the distribution is used to count the number of times an event happens, and an event cannot be halved.
- The events are independent of each other. So the occurrence of one event does not affect the probability of another event happening.
- A Poisson distribution is very interesting because when the mean (λ) increases, the distribution of values becomes more symmetric and, thus, more bell-shaped. When the mean is large enough, we can apply the Central Limit Theorem to approximate the Poisson distribution by a normal distribution N(λ,λ). Here, the approximation improves as the mean increases.
Importance of Poisson Distribution
Poisson distribution is important because of its wide application. Here are few other reasons why it is such a prominent tool in mathematics:
- Poisson distribution is ideal for modeling situations where events occur randomly and independently. One such example is the number of emails you receive per hour.
- It's very useful in predicting the probability of rare events in a particular time period. Such as the number of earthquakes happening in a year.
- Insurance companies use Poisson Distribution to assess risks or accidents happening in a specific time frame.
Tips and Tricks to Master Poisson Distribution
It can be quite confusing when learning about Poisson distribution. Here are a few tips and tricks students can use to master Poisson distribution.
- Remember the formula: Learning the formula is very important when learning Poisson distribution. P(x) =(X = k) = e-λ λkk!
- The value of mean and variance will be the same. This is a very significant and unique property to remember. So if the λ = 7, then the mean and variance = 7.
- Know when to use the Poisson Distribution. We use it when we want to find out how many times an event occurs. So if you want to figure out how many calls you receive per hour, we use Poisson Distribution.
Real-World Applications on Poisson Distribution
Poisson distribution is widely used in the real world. Here are a few examples where we use Poisson Distribution.
- Call Centers: To figure out the number of customer calls received at a call center per minute, we use a Poisson distribution. This helps managers allocate the resources efficiently.
- Natural Disasters: Natural disasters like earthquakes and their frequency of occurrence in a particular place can be analyzed using Poisson distribution.
- Detecting Email Spams: By modeling the frequency of incoming emails per day, a Poisson distribution can be used to enhance filtering algorithms.This will allow users to analyze and predict email traffic patterns.
Common Mistakes and How to Avoid Them in 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:
Mistake 1
Incorrectly interpreting λ
Remember that λ is the average rate of occurrences and not the probability. λ remains constant for the entire issue.
Mistake 2
Not calculating the factorials properly
When solving for k! Use a calculator or factorial tables. Students need to remember that when solving for k! They need to be careful as it is possible to make mistakes. Double-check your calculations for small or larger numbers.
Mistake 3
Use the constant e incorrectly
Students may forget the value of e (2.718) and it should be raised to the power of -λ. When solving, ensure you apply the negative sign correctly and use the constant properly.
Mistake 4
Speeding through the formula substitution
When substituting values in the formula, students may rush and substitute the values. This can lead to mistakes because sometimes the conditions might not have been met. So read the issue carefully and see if all the conditions (independent events, fixed interval, constant rate) are met.
Mistake 5
Overlooking the fact that only whole numbers are valid
Students may overlook the property that Poisson distribution is discrete, which means that whole numbers are valid. We cannot use non-integer values such as 4.7.
Explore More data
![Important Math Links Icon]()
Previous to Poisson Distribution
![Important Math Links Icon]()
Next to Poisson Distribution
Solved Examples on Poisson Distribution
Problem 1
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.
Explanation
\(P(x) = (X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
P(3) = (X = k) = \(\frac{e^{-2} \times 2^3}{3!} \) = \(\frac{8e^{-2}}{3!} \)≈ \(0.1353 \times 8 \over 6 \)≈\(1.0824\over6\)≈ 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 23 = 8 and 3! = 6.
- Use e-2 ≈ 0.1353 and calculate the final probability.
Problem 2
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
Explanation
For 2 hours, λ = 4 × 2 = 8.
\(P(x) = (X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
P(5) = (X = k) = \(\frac{8^5 e^{-8}}{5!}\)≈ \(0.000335 \times 32768 \over 120\)≈\(11.0 \over 120\)≈ 0.0917
- First convert the rate of calls: 4 calls per hour × 2 hours = 8 calls in 2 hours.
- Substitute k = 5
- Calculate 85 = 32768 and 5! = 120.
- Use e-8 ≈ 0.000335 (where e is 2.718) and calculate the final probability.
Problem 3
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.
Explanation
Calculate λ = 1000 × 0.005 = 5.
Then,
\(P(x) = (X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
P(3) = (X = k) = \(\frac{e^{-5} \times 5^3}{3!} \)≈\(0.00674 \times 125 \over6\)≈\(0.8425 \over6\)≈ 0.1404
Problem 4
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
Explanation
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.
Problem 5
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.
Explanation
\(P(x) = (X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
P(5) = (X = k) = \(\frac{e^{-5} \times 5^4}{4!} \)≈ \( \frac{e^{-3} \times 3^5}{5!} \)≈ \(0.0498 \times243 \over120\)≈ \(12.1014 \over120\)≈ 0.1008
1.When should Poisson Distribution be used?
2.Why is Poisson Distribution said to be discrete?
3.How do I find the probability using Poisson Distribution?
4.If the events are dependent on each other, can we use Poisson distribution?
5.What tools can we use to help in calculating Poisson distribution?
Jaipreet Kour Wazir
About the Author
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
Fun Fact
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!
