13. The Poisson Probability Distribution
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The Poisson Distribution was developed by the French mathematician Simeon Denis Poisson in 1837.
The Poisson random variable satisfies the following conditions:
The number of successes in two disjoint time intervals is independent.
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 lowercase 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
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) = σ^{2} = 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
Some policies

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

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(x_{0})
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)`
`=1P(x_0)`
`=14.9787xx10^2`
`=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!)`
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`
Example 4
Vehicles pass through a junction on a busy road at an average rate of `300` per hour.

Find the probability that none passes in a given minute.

What is the expected number passing in two minutes?

Find the probability that this expected number actually pass through in a given twominute 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:
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 viceversa) under certain circumstances.