The Poisson Distribution


  • The probability function of the poisson distribution is P(X\quad =\quad x)\quad =\quad \frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! }
  • Both the mean and variance the same in poisson distribution.

When calculating poisson distribution the first thing that we have to keep in mind is the if the random variable is a discrete variable. If however, your variable is a continuous variable e.g it ranges from 1<x<2 then poisson distribution cannot be applied.

The possible values of the poisson distribution are the non-negative integers 0,1,2… The probability function of the poisson distribution is:

P(X\quad =\quad x)\quad =\quad \frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! }

Where \lambda (pronounced lamda) is the mean, which is calculated as [n.p]

Where n is the total number of trials and

P is the successful probability

Note: Always remember that both x and p will always stay associated with each other e.g if x is a success trial then p will also be a success trial.

Both the mean and variance the same in poisson distribution

x\quad \sim \quad poisson(\lambda )

\lambda in this case is considered as the parameter of the distribution. We are already familiar with the idea that poisson distribution occurs when there are discrete events in a continuous time.

E.g we can say that if you have 4 phone calls in one hour, then the number of phone calls at any given time in that hour has poisson distribution with parameter 4.

The mean of the poisson distribution would be:

E(X)\quad =\quad \lambda

The variance of the poisson distribution would be:

var(X)\quad =\quad \lambda

Poisson distribution has many properties like:

  1. The trials are independent
  2. The events cannot occur simultaneously
  3. Events are random and unpredictable
  4. The poisson distribution provides an estimation for binomial distribution.

Sum of poissons

Consider the sum of two independent random variables X and Y with parameters L and M. Then the distribution of their sum would be written as:

X\quad \sim \quad po(L)

Y\quad \sim \quad Po(M)


X\quad +\quad Y\quad \sim \quad po(L\quad +\quad M)


Q. A hospital board receives an average of 4 emergency calls in 10 minutes. What is the probability that there are at most 2 emergency calls?


Step #1

We will first find the \lambda and x.

\lambda also known as the mean or average or expectation, has been provided in the question.

\lambda = 4

X\quad =\quad \le \quad 2 its less than equal to 2 since the question says at most. Which means, maximum 2 not more than that.

Step #2

We will now plug the values into the poisson distribution formula for:

P[ \le 2] = P(X=0) + P(X=1)+(PX=2)

The mean will remain same throughout, however, the value of x will change (0, 1, 2)

P(X\quad =\quad 0)\quad =\quad \frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } \quad =\quad \frac { { e }^{ -4 }{ 4 }^{ 0 } }{ 0! } \quad =\quad 0.0183

P(X\quad =\quad 1)\quad =\quad \quad \frac { { e }^{ -4 }{ 4 }^{ 1 } }{ 1! } \quad =\quad 0.7328

P(X\quad =\quad 2)\quad =\quad \quad \frac { { e }^{ -4 }{ 4 }^{ 2 } }{ 2! } \quad =\quad 0.1465

Note: the value of e will remain constant e\quad \approx \quad 2.71828.


Add all the answers together to get the final answer.

P[latex]\le \quad 2[/latex] = 0.0183 + 0.7328 +0.1465

= 0.8976

Example #2

Q. Births in a hospital occur randomly at an average rate of 1.8 births per hour. What is the probability of observing 4 births in a given hour at the hospital?


Calculate both \lambda and X

\lambda = 1.8

X = P(X = 4)

This question does not state anything else but to find the probability of 4 births in an hour, thus we will just simply calculate P(X = 4)


We will now use the formula of poisson distribution to calculate the answer.

P(X\quad =\quad 4)\quad =\quad \frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } \quad =\quad \frac { { e }^{ -1.8 }{ 1.8 }^{ 4 } }{ 4! } \quad =\quad 0.0723

We can thus say, the probability of observing 4 births in a given hour at the hospital is 0.0723