PoissonDistribution - Maple Help
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The Poisson Distribution

Main Concept

The Poisson distribution is a discrete probability distribution that gives the probability of a given number of events k occurring in a fixed interval of time. In a Poisson distribution, events are assumed to occur with a known constant rate, λ, independent of the time since the last event.


In order for an event to be described by the Poisson distribution, there are three conditions in which the event must hold:


Independence: The events occur in disjoint intervals (non-overlapping).


Individuality: Two or more events cannot occur simultaneously.


Uniformity: Each event occurs at a constant rate.


If all three conditions are met, it can be said that the event occurs according to a Poisson Process.

If a random variable X follows a poisson distribution, the probability of k events occurring in a given interval of time is given by the following probability mass function:

fk = PX=k= λkⅇλk! ,

for k = 0,1,2, ... ,


Where λ is the constant rate or intensity at which the event occurs and k is the number of events.


The cumulative distribution function is defined as:


fk = PXk = ⅇλ i=0kλii!,

for k = 0,1,2, ... 


Note that the probability function and cumulative distribution function for the poisson distribution are only defined for integer values for k and as such, there is no continual curve which can be drawn through the point.  The gray lines are only for illustrating the shape of each function over the interval.


If l is the rate or intensity at which the Poisson Process occurs at and k is the number of events then:



The probability mass function


ⅇλ i=0kλii!

The cumulative distribution function

Mean E(X)


The expected value of a random variable

Variance Var(X)


Represented by the symbol σ2, representing how much variation or spread exists from the mean value




Suppose you are testing a new software and a bug causes errors randomly at a constant rate of once every 5 hours. What is the probability that four bugs will occur in the next 5 hours?


Let rate or intensity be  l = 1 per 5 hours and number of events x be 4. 

fX =4 = 14e14!= 0.0153

Therefore, the probability that four bugs will occur in the next 5 hours is 0.0153.



Change the number of successes k, and the intensity of the event l, to observe the change in the probability of k events occurring:

rate of event (l) =

# of success (k) =

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