Poisson Approximation - Definition, Solved Examples in Probability
The Poisson distribution is an important probability distribution which is the resultant of the Poisson experiment. The Poisson distribution was discovered by Simeon Denis Poisson who was a French mathematician and was published in his book in 1837. It is said to be a discrete kind of probability distribution which represents the probability of certain number of events that occur in a fixed interval and these events have an average rate of occurring.
Poisson experiment has following features:

1) The result of a Poisson experiment must be notated in the form of success or failure.

2) In a specified interval or trials, the average number of successes must be known.

3) For a very small region, the probability of occurring a success should be zero.

4) The probability of obtaining a success should be proportional to the size of given region.
In this article, we shall go ahead and learn about Poisson approximation in detail.


A random variable Z with parameter $\mu$ > 0 is said to be in Poisson distribution if:

$P (Z = k)$ = $\frac{e^{- \mu} \mu^{k}}{k!}$, for k = 0, 1, …

A binomial distribution is represented by:

Bin (n, s) = $C^{n}_{k} s^{k} (1 - s)^{(n - k)}$

If n is very large and ‘k’ is small relative to s then, it is equal to:

$\frac{\mu^{k}}{k! (1 - \frac{\mu}{n})^n}$

$\approx$ $\frac{\mu^{k}}{k! . e^(- \mu)}$ as n is large.

So we see that with some conditions the Poisson appears as an approximation to Binomial distribution too.

With large value of ‘n’, it is very difficult to calculate the Binomial distribution. So when ‘n’ approaches to infinity (n -> 8), ‘s’ approaches to zero (s -> 0), then n . s = $\mu$(a constant). In this case, Binomial (n, s) approaches to Poisson ($\mu$), that is

$\frac{n !}{k ! (n – k) !} s^{k} (1 - s)^{(n – k)}$ $\Rightarrow$ $\frac{e^{- \mu} \mu^{k}}{k!}$

For Example: suppose 5 out of 50,000 pens are defective. If Y represent the count of defective pens in a batch of 1,00,000, then the probability of getting at least 6 of them defective can only be deduced not by using Binomial distribution, but by the Poisson approximation of the Binomial in this case. Here n = 1,00,000, s = 0.0001, n . s = 10. So for \mu = n . s = 10

Chances of getting at least 6 defective 

= 1 - chances of getting less than 6 defective

= 1 - [ P ( no defective) + P (one defective) +  …. + P (five defective)]


Have a closer look at an example of Poisson approximation:

Example: The birth rate of a particular bacteria is noted to be 1.8 per microsecond. What is the probability of the following :
i) Occurring 4 births in a microsecond ?

ii) Occurring more than or equal to 2 births in a microsecond ?

Average rate $\mu$ = 1.8

We have the following formula for Poisson approximation

$P (Z = k)$ = $\frac{e^{- \mu} \mu^{k}}{k!}$

i) Determination of probability of occurring 4 births would be given by:

P(X = 4) = $\frac{e^{−1.8} 1.8^{4}}{4!}$

$\frac{0.1653 \times 10.4976}{24}$

= 0.0723

ii) P(X $\geq$ 2) = P(X = 2) + P(X = 3) + ...

P(X $\geq$ 2) = 1 - P(X < 2)

= 1 - [P(X = 0) + P(X = 1)]

= 1  [$\frac{e^{-1.8}1.8^{0}}{0!}$ + $\frac{e^{-1.8} 1.8^{1}}{1!}$]

= 1 - (0.16529 + 0.29753)

= 0.537