Poisson Distribution Formula
A Poisson experiment is a statistical experiment and a theoretical discrete probability that classifies the experiment into two categories, success or failure. Poisson distribution is a limiting process of the binomial distribution. The Poisson distribution formula is very useful in situations where discrete events occur in a continuous manner.
Let's see the following properties of a Poisson model:
 The event or success is something that can be counted in whole numbers.
 The probability of having success in a time interval is independent of any of its previous occurrence.
 The average frequency of successes in a unit time interval is known.
 The probability of more than one success in a unit time is very low.
Let us understand the Poisson distribution formula using solved examples.
What Is Poisson Distribution Formula?
A Poisson random variable “x” is used to define the number of successes in the experiment. This distribution generally occurs when there are events that do not occur as the outcomes of a definite number of outcomes.
Poisson distribution is used under certain conditions:
 The number of trials, n, tends to infinity
 Probability of success, p, tends to zero
 np = 1 is finite
For poison random variable, \(x = 0,1,2, \ldots \infty \), the Poisson distribution formula is given by:
\(f\left( x \right) = P\left( {X = x} \right) = \frac{{e^{  \lambda } \lambda ^x }}{{x!}}\)
where,
 e is the base of the logarithm
 x is a Poisson random variable
 λ is an average rate of value
In shorthand notation, it is represented as
\(X \sim P\left( \lambda \right)\)
Poisson Distribution Formula is used when the independent events occurring at a constant rate within the given interval of time are provided.
Let us understand the Poisson distribution formula using solved examples.
Solved Examples Using Poisson Distribution Formula

Example 1: If the random variable X follows a Poisson distribution with a mean of 3.4, find P(X = 6).
Solution:
Given that; X ~ Po(3.4)
Find P(X = 6).
Now, Using the Poisson distribution formula.
\( P( {X = 6} ) = \frac{{e^{  \lambda } \lambda ^6 }}{{6!}} \\
= \frac{{e^{  3.4} ( {3.4} )^6 }}{{6!}} \\
= 0.071604409 \\
= 0.072 \)
Answer: \(P(X = 6)= 0.072\) 
Example 2: A factory produces nails and packs them in boxes of 200. If the probability that a nail is substandard is 0.006, find the probability that a box selected at random contains at most two nails that are substandard. Use the Poisson distribution formula.
Solution:
If X is the number of substandard nails in a box of 200, then
\(X \sim B\left( {200,0.006} \right)\)
Since, n is large and p is small, the Poisson approximation can be used.
The appropriate value of \(\lambda\) is given by
\(\lambda = np = 200 \times 0.006 = 1.2\)
So, \(X \sim Po\left( {1.2} \right)\) and
\(
P\left( {X \le 2} \right) = e^{  1.2} + e^{  1.2} \times 1.2 + \frac{{e^{  1.2} \left( {1.2} \right)^2 }}{{2!}} \\ = 2.92e^{  1.2} \\ = 0.8795 \\
\)
Answer: The probability that a box selected at random contains at most two nails that are substandard is 0.8795