# 6.8: Poisson Distribution


Random variables that can be thought of as “How many occurrences per time period”, or “How many occurrences per region” have many practical applications. For example:

The number of strong earthquakes per year  in California.

The number of customers per hour at a restaurant.

The number of accidents per week at a manufacturing plant.

The number of errors per page in a manuscript.

If the rate is constant, these random variables will follow a Poisson distribution.

The Poisson Distribution is actually derived from a Binomial Distribution in which the sample size $$n$$ gets very large and the probability of success $$p$$ is very small. A good example of this is the Powerball Lottery.

Example: Powerball Lottery

The odds of winning the Powerball Lottery jackpot with a single ticket are 292,000,000 to 1. Suppose the jackpot gets large and 292,000,000 tickets are sold.

Solution

Let $$X$$ = Number of jackpot winning tickets sold.

Under the Binomial distribution, $$n=292,000,000$$ and $$p = 1/292,000,000$$. Note that $$p$$ is very close to zero, so $$1‐p$$ is very close to 1.

$$\mu=n p=1$$

$$\sigma^{2}=n p(1-p) \approx n p=\mu=1$$

The number of winners can be modeled by the Poisson Distribution, in which the single parameter $$\(\mu$$ is the expected number of winners; in this case $$\mu=1$$. There could theoretically be millions of winners, so the possible values of the Poisson is designed so there is no theoretical limit for the value of $$X$$ (although there are practical limits in real life problems).

The important features of the Poisson Distribution are shown here:

Poisson Probability Distribution (parameter= $$\mu$$)

$$\mu$$ = expected occurrences per given time period or region. This rate must be constant.

$$X$$ = number of occurrence per given time period or region Possible values of $$X$$ {0, 1, 2, …} (no upper limit)

$$\sigma^{2}=\mu$$

$$\sigma=\sqrt{\mu}$$

$$P(x)=\dfrac{e^{-\mu} \mu^{x}}{x !}$$

Example: Continuation of Powerball Lottery

Find the probability of no jackpot winners.

$$P(0)=\dfrac{e^{-1} 2^{0}}{0 !}=0.368$$

Find the probability of at least one jackpot winner. The answer calculated directly is an infinite sum, so instead use the Rule of Complement

$$P(X \geq 1)=P(1)+P(2)+\cdots$$

$$P(X \geq 1)=1-P(0)=1-\dfrac{e^{-1} 2^{0}}{0 !}=0.632$$

There is a 63.2% chance that at least one winning ticket is sold.

Example: Earthquakes

Earthquakes of Richter magnitude 3 or greater occur on a certain fault at a rate of two times per every year. Assume this rate is constant.

Solution

Find the probability of at least one earthquake of RM 3 or greater in the next year.

$$\mu=2$$ per year.

$$P(X \geq 1)=1-P(0)=1-\dfrac{e^{-2} 2^{0}}{0 !}=0.865$$

Find the probability of exactly 6 earthquakes of RM 3 or greater in the next 2 years.

When determining the parameter m for the Poisson Distribution, make sure that the expected value is over the time period or region given in the problem. Since these earthquakes occur at a rate of 2 per year, we would expect 4 earthquakes in 2 years.

$$\mu$$= (2 per year)( 2 years) = 4

$$P(X=6)=\dfrac{e^{-6} 4^{0}}{6 !}=0.104$$

Counting methods that are modeled by random variables that follow a Poisson Distribution are also called a Poisson Process.

This page titled 6.8: Poisson Distribution is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Maurice A. Geraghty via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.