# Poisson Probability distribution Examples and Questions

Poisson probability distribution is used in situations where events occur randomly and independently a number of times on average during an interval of time or space. The random variable \( X \) associated with a Poisson process is discrete and therefore the Poisson distribution is discrete.

## Poisson Process Examples and Formula

Example 1

These are examples of events that may be described as Poisson processes:

- My computer crashes on average once every 4 months.

- Hospital emergencies receive on average 5 very serious cases every 24 hours.

- The number of cars passing through a point, on a small road, is on average 4 cars every 30 minutes.

- I receive on average 10 e-mails every 2 hours.

- Customers make on average 10 calls every hour to the customer help center

Conditions for a Poisson distribution are

1) Events are discrete, random and independent of each other.

2) The average number of times of occurrence of the event is constant over the same period of time.

3) Probabilities of occurrence of event over fixed intervals of time are equal.

4) Two events cannot occur at the same time; they are mutually exclusive.

In a Poisson distribution, if an event happens an average \( \lambda \) times over a period \( T \) of time or space, the probability that it will happen \( x \) times over a period of time \( T \) is given by

\[ P(X = x) = \dfrac{e^{-\lambda}\lambda^x}{x!} \] where \( e \approx 2.7182818\) is the base of the natural logarithm, \( x! \) is the factorial of \( x \) defined as \( x! = 1 \times 2 \times 3 ... \times (x - 1) \times x \) and \( x = 0, 1, 2, ....\infty \)

Below we show the graphs of \( P(X) \) for several values of the average \( \lambda \) and we note that the probability is maximum for \( x \) close to the average \( \lambda \) and decreases as \( x \) takes larger values which makes sense.

## Compare Binomial and Poisson Distributions

A binomial distribution has two parameters: the number of trials \( n \) and the probability of success \( p \) at each trial while a Poisson distribution has one parameter which is the average number of times \( \lambda \) that the event occur over a fixed period of time.In the binomial distribution \( x \) is an integer taking values over the interval \( [0 , n] \) , while in the Poisson distribution \( x \) is an integer taking values over the interval \( [0 , \infty) \)

## Poisson distribution Examples with Detailed SolutionsThe best way to explain the formula for the Poisson distribution is to solve the following example.
Example 2
Example 3
Example 4
Example 5
a) one goal in a given match b) at least one goal in a given match Solution to Example 5a) We first calculate the mean \( \lambda\) \( \lambda = \dfrac{\Sigma f \cdot x}{\Sigma f} = \dfrac{12 \cdot 0 + 15 \cdot 1 + 6 \cdot 2 + 2 \cdot 3 }{ 12 + 15 + 6 + 2} \approx 0.94 \) The probability that he will score one goal in a match is given by the Poisson probability formula \( P(X = 1) = \dfrac{e^{-\lambda}\lambda^x}{x!} = \dfrac{e^{- 0.94} 0.94^1}{1!} = 0.36719 \) b) Al least one goal means 1 or 2 or 3 or 4 .... goals \( P(X \ge 1) = P(X=1 \; or \; X=2 \; or \; X=3 ... ) \) Using the complement \( = 1 - P(X = 0) \) Substitute by formulas \( = 1 - \dfrac{e^{-0.94}0.94^0}{0!} \) \( = 1 - 0.39062 \) \( = 0.60938 \)
Example 6
a) no defective item is returned on a given day b) three or more defective items are returned on a given day Solution to Example 6a) We first calculate the mean \( \lambda\) \( \lambda = \dfrac{\Sigma f \cdot x}{\Sigma f} = \dfrac{50 \cdot 0 + 20 \cdot 1 + 15 \cdot 2 + 10 \cdot 3 + 5 \cdot 4 }{ 50 + 20 + 15 + 10 + 5} = 1 \) The probability that no defective item is returned is given by the Poisson probability formula \( P(X = 0) = \dfrac{e^{-\lambda}\lambda^x}{x!} = \dfrac{e^{- 1} 1^0}{0!} = 0.36787 \) b) At least three or more defective items are returned means 3 or 4 or 5 .... itmes or \( x \ge 3 \) \( P(X \ge 3) = P(X=3 \; or \; X=4 \; or \; X=5 ... ) \) Using the complement \( = 1 - P(X=0 \; or \; X=1 \; or \; X=2) \) Use the sum formula of probabilities and Poisson by formula \( = 1 - (\dfrac{e^{-1} 1^0}{0!} + \dfrac{e^{-1} 1^1}{1!} + \dfrac{e^{- 1} 1^2}{2!}) \) \( = 1 - 0.91969 \) \( = 0.08031 \) ## More References and linksPoisson Probability Distribution CalculatorBinomial Probabilities Examples and Questions addition rule of probabilities multiplication rule of probabilities probability questions classical formula for probability mutually exclusive events Introduction to Probabilities sample space event elementary statistics and probabilities. Home Page |