Poisson Distribution

The Poisson Distribution gives the probability of a specific number of independent events occurring in a fixed interval when the average rate is known.

P(X=k)=λkeλk!P(X = k) = \frac{\lambda^{k}e^{-\lambda}}{k!}

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What each symbol means

What Poisson Distribution takes
XX
kk
λ\lambda
Poisson Distribution
SymbolMeaning
XXThe random variable representing the number of events that occur in the fixed interval.
kkThe specific number of events for which you are calculating the probability; must be a non-negative integer.
λ\lambdaThe average (expected) number of events occurring in the fixed interval; must be positive.

When to use it

Use Poisson when you know the average rate of independent events occurring in a fixed interval.

Level

Usually taught in: Statistics

Worked examples

1. Find the probability of exactly 2 customer inquiries

Problem

A bookstore receives an average of 4 customer inquiries per hour. What is the probability that exactly 2 inquiries arrive in the next hour?
  1. P(X=2)=42e42!P(X = 2) = \frac{4^{2}e^{-4}}{2!}

    Substitute λ = 4 and k = 2 into the Poisson formula.

  2. P(X=2)=16e42P(X = 2) = \frac{16 \cdot e^{-4}}{2}

    Calculate 42=164^2 = 16 and 2!=22! = 2.

  3. P(X=2)=160.01832P(X = 2) = \frac{16 \cdot 0.0183}{2}

    Use e40.0183e^{-4} \approx 0.0183.

  4. P(X=2)=0.29282P(X = 2) = \frac{0.2928}{2}

    Multiply 16×0.0183=0.292816 \times 0.0183 = 0.2928.

  5. P(X=2)0.1465P(X = 2) \approx 0.1465

    Divide to get the final probability, or about 14.65%.

Answer: P(X=2)0.1465 or 14.65%P(X = 2) \approx 0.1465 \text{ or } 14.65\%

This uses the Poisson formula directly because the inquiries arrive independently at a known average rate per unit time, making Poisson the natural choice.

2. Find the probability of zero server errors in a day

Problem

A server receives an average of 2.5 error messages per day. Find the probability of exactly 0 errors on a given day.
  1. P(X=0)=2.50e2.50!P(X = 0) = \frac{2.5^{0}e^{-2.5}}{0!}

    Substitute λ = 2.5 and k = 0 into the Poisson formula.

  2. P(X=0)=1e2.51P(X = 0) = \frac{1 \cdot e^{-2.5}}{1}

    Use 2.50=12.5^0 = 1 and 0!=10! = 1.

  3. P(X=0)=e2.5P(X = 0) = e^{-2.5}

    Simplify to get just e2.5e^{-2.5}.

  4. P(X=0)0.0821P(X = 0) \approx 0.0821

    Use e2.50.0821e^{-2.5} \approx 0.0821.

Answer: P(X=0)0.0821 or 8.21%P(X = 0) \approx 0.0821 \text{ or } 8.21\%

When calculating the probability of zero events, k0=1k^0 = 1 and 0!=10! = 1, so the formula simplifies nicely to just eλe^{-\lambda}.

3. Find the probability of exactly 5 radioactive decay events

Problem

A chemistry lab measures radioactive decay from a sample. Historically, the sample produces an average of 6 decay events per 10-second interval. What is the probability of observing exactly 5 decay events in the next 10 seconds?
  1. P(X=5)=65e65!P(X = 5) = \frac{6^{5}e^{-6}}{5!}

    Substitute λ = 6 and k = 5 into the Poisson formula.

  2. 65=7776 and 5!=1206^{5} = 7776 \text{ and } 5! = 120

    Calculate 65=77766^5 = 7776 and 5!=1205! = 120.

  3. P(X=5)=7776e6120P(X = 5) = \frac{7776 \cdot e^{-6}}{120}

    Substitute the computed values and use e60.00248e^{-6} \approx 0.00248.

  4. P(X=5)=77760.00248120=19.281200.1607P(X = 5) = \frac{7776 \cdot 0.00248}{120} = \frac{19.28}{120} \approx 0.1607

    Multiply and divide: 7776×0.0024819.287776 \times 0.00248 \approx 19.28, then divide by 120.

Answer: P(X=5)0.1607 or 16.07%P(X = 5) \approx 0.1607 \text{ or } 16.07\%

Radioactive decay is a classic Poisson process: events occur independently, the average rate is constant, and we want the probability of a specific count, making this formula ideal.

Common mistakes

Where Poisson Distribution usually goes wrong
Answer came out wrong
Writing P(X=k)=λkeλk!λP(X = k) = \frac{\lambda^k e^{-\lambda}}{k! \cdot \lambda}
The denominator is only k!k!, giving P(X=k)=λkeλk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}.
Thinking that P(X=k)=0P(X = k) = 0 whenever k>λk > \lambda, assuming you cannot observe more events than the average.
Calculate P(X=k)P(X = k) normally—the probability will be small but non-zero whenever k1k \geq 1.
Confusing λ (the average rate) with k (the specific count of events being evaluated).
Remember that λ is known upfront from the data or problem statement, and you plug in different values of k to get probabilities for each possible outcome.
The mistakeWhy it is wrongThe fix
Writing P(X=k)=λkeλk!λP(X = k) = \frac{\lambda^k e^{-\lambda}}{k! \cdot \lambda}Dividing by λ in the denominator is incorrect; only k!k! belongs there.The denominator is only k!k!, giving P(X=k)=λkeλk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}.
Thinking that P(X=k)=0P(X = k) = 0 whenever k>λk > \lambda, assuming you cannot observe more events than the average.It is completely possible to observe more events than the average; Poisson allows this with non-zero probability.Calculate P(X=k)P(X = k) normally—the probability will be small but non-zero whenever k1k \geq 1.
Confusing λ (the average rate) with k (the specific count of events being evaluated).These represent different things: λ is a fixed property of the process, while k is the outcome you are asking about.Remember that λ is known upfront from the data or problem statement, and you plug in different values of k to get probabilities for each possible outcome.

Tips and when to use something else

  • Poisson works best when events are rare or moderate in frequency; if you have a fixed number of trials with two outcomes, use Binomial Probability instead.
  • The expected value of a Poisson distribution is λ, and the variance is also λ—one of Poisson's special properties.
  • When λ is large (typically > 20), the Poisson distribution approaches a normal distribution.
  • For quick approximations, remember that P(X=0)=eλP(X = 0) = e^{-\lambda}, so single-event probabilities often simplify nicely.

Frequently asked questions

When should I use Poisson instead of Binomial?
Use Poisson when you have a large number of trials but a small probability of success in each trial (so npλnp \approx \lambda is moderate), and no fixed number of trials. Binomial requires a fixed n. Poisson is also simpler computationally when n is large and p is small.
Can λ be a non-integer or a decimal?
Yes—λ can be any positive number. It represents the average rate, which need not be a whole number. For example, if a process averages 2.5 events per hour, you use λ = 2.5.
What does the e in the formula mean?
The e is Euler's number, approximately 2.71828, the base of natural logarithms. The term eλe^{-\lambda} appears because it describes the probability of zero events occurring—a core part of the Poisson model.
Is there a simpler way to calculate Poisson probabilities without knowing e?
Most calculators and statistical software have Poisson functions built in; you can often just input λ and k without manually computing eλe^{-\lambda}. For hand calculation, a table of eλe^{-\lambda} values or a scientific calculator is essential.

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Reviewed 2026-09-18