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13. The Poisson Probability Distribution

The Poisson Distribution was developed by the French mathematician Simeon Denis Poisson in 1837.

The Poisson random variable satisfies the following conditions:

  1. The number of successes in two disjoint time intervals is independent.

  2. The probability of a success during a small time interval is proportional to the entire length of the time interval.

Apart from disjoint time intervals, the Poisson random variable also applies to disjoint regions of space.

Applications

  • the number of deaths by horse kicking in the Prussian army (first application)
  • birth defects and genetic mutations
  • rare diseases (like Leukemia, but not AIDS because it is infectious and so not independent) - especially in legal cases
  • car accidents
  • traffic flow and ideal gap distance
  • number of typing errors on a page
  • hairs found in McDonald's hamburgers
  • spread of an endangered animal in Africa
  • failure of a machine in one month

Notation

We use upper case variables (like X and Z) to denote random variables, and lower-case letters (like x and z) to denote specific values of those variables.

The probability distribution of a Poisson random variable X representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:

`P(X)=(e^{-mu} mu^x)/(x!)`

where

`x = 0, 1, 2, 3...`

`e = 2.71828` (but use your calculator's e button)

`μ =` mean number of successes in the given time interval or region of space

Continues below

Mean and Variance of Poisson Distribution

If μ is the average number of successes occurring in a given time interval or region in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to μ.

E(X) = μ

and

V(X) = σ2 = μ

Note: In a Poisson distribution, only one parameter, μ is needed to determine the probability of an event.

Example 1

A life insurance salesman sells on the average `3` life insurance policies per week. Use Poisson's law to calculate the probability that in a given week he will sell

  1. Some policies

  2. `2` or more policies but less than `5` policies.

  3. Assuming that there are `5` working days per week, what is the probability that in a given day he will sell one policy?

Answer

Here, μ = 3

(a) "Some policies" means "1 or more policies". We can work this out by finding 1 minus the "zero policies" probability:

P(X > 0) = 1 − P(x0)

Now `P(X)=(e^(-mu)mu^x)/(x!)` so `P(x_0)=(e^-3 3^0)/(0!)=4.9787xx10^-2`

Therefore the probability of `1` or more policies is given by:

`"Probability"=P(X>=0)`

`=1-P(x_0)`

`=1-4.9787xx10^-2`

`=0.95021`

(b) The probability of selling 2 or more, but less than 5 policies is:

`P(2<=X<5)`

`=P(x_2)+P(x_3)+P(x_4)`

`=(e^-3 3^2)/(2!)+(e^-3 3^3)/(3!)+(e^-3 3^4)/(4!)`

`=0.61611`

(c) Average number of policies sold per day: `3/5=0.6`

So on a given day, `P(X)=(e^-0.6(0.6)^1)/(1!)=0.32929`

Example 2

Twenty sheets of aluminum alloy were examined for surface flaws. The frequency of the number of sheets with a given number of flaws per sheet was as follows:

Number of flaws Frequency
`0` `4`
`1` `3`
`2` `5`
`3` `2`
`4` `4`
`5` `1`
`6` `1`

What is the probability of finding a sheet chosen at random which contains 3 or more surface flaws?

Answer

The total number of flaws is given by:

`(0 × 4) + (1 × 3)` `+ (2 × 5) + (3 × 2)` `+ (4 × 4) + (5 × 1)` `+ (6 × 1)` `= 46`

So the average number of flaws for the 20 sheets is given by:

`mu=46/20=2.3`

The required probability is:

`"Probability"=P(X>=3)`

`=1-(P(x_0)+P(x_1)+P(x_2))`

`=1-((e^-2.3 2.3^0)/(0!)+(e^-2.3 2.3^1)/(1!)+(e^-2.3 2.3^2)/(2!))`

`=0.40396`

Histogram of Probabilities

We can see the predicted probabilities for each of "No flaws", "`1` flaw", "`2` flaws", etc on this histogram.

The histogram was obtained by graphing the following function for integer values of x only.

`(e^-2.3 2.3^x)/(x!)`

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Example 3

If electricity power failures occur according to a Poisson distribution with an average of `3` failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.

Answer

The average number of failures per week is: `mu=3/20=0.15`

"Not more than one failure" means we need to include the probabilities for "`0` failures" plus "`1` failure".

`P(x_0)+P(x_1)` `=(e^-0.15 0.15^0)/(0!)+(e^-0.15 0.15^1)/(1!)` `=0.98981`

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Example 4

Vehicles pass through a junction on a busy road at an average rate of `300` per hour.

  1. Find the probability that none passes in a given minute.

  2. What is the expected number passing in two minutes?

  3. Find the probability that this expected number actually pass through in a given two-minute period.

Answer

The average number of cars per minute is: `mu=300/60=5`

(a) `P(x_0)=(e^-5 5^0)/(0!)=6.7379xx10^-3`

(b) Expected number each 2 minutes = E(X) = 5 × 2 = 10

(c) Now, with μ = 10, we have: `P(x_10)=(e^-10 10^10)/(10!)=0.12511`

Histogram of Probabilities

Based on the function

`P(X)=(e^-10 10^x)/(x!)`

we can plot a histogram of the probabilities for the number of cars for each 2 minute period:

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Example 5

A company makes electric motors. The probability an electric motor is defective is `0.01`. What is the probability that a sample of `300` electric motors will contain exactly `5` defective motors?

Answer

The average number of defectives in 300 motors is μ = 0.01 × 300 = 3

The probability of getting `5` defectives is:

`P(X)=(e^-3 3^5)/(5!)=0.10082`

NOTE: This problem looks similar to a binomial distribution problem, that we met in the last section.

If we do it using binomial, with `n = 300`, `x = 5`, `p = 0.01` and `q = 0.99`, we get:

P(X = 5) = C(300,5)(0.01)5(0.99)295 = 0.10099

We see that the result is very similar. We can use binomial distribution to approximate Poisson distribution (and vice-versa) under certain circumstances.

Histogram of Probabilities

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