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Last updated on October 4, 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) = (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.
Knowing about the key properties will help us utilize Poisson distribution better. Here are a few important properties:
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 23 = 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 85 = 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) = \(\frac{e^{-5} \times 5^3}{3!} \)≈\(0.00674 \times 125 \over6\)≈\(0.8425 \over6\)≈ 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) = \(\frac{e^{-5} \times 5^4}{4!} \)≈ \( \frac{e^{-3} \times 3^5}{5!} \)≈ \(0.0498 \times243 \over120\)≈ \(12.1014 \over120\)≈ 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!