Poisson Distribution: Overview, Questions, Preparation

Poisson distribution, a concept first introduced by French mathematician Siméon Denis Poisson in 1930. It is another probability distribution that expresses the probability of an event taking place. The unique factor is that the individual knows the said rate of occurrence event. It is considered to be a discrete function. It implies that one can measure the event’s probability as occurring and not occurring. Poisson distribution occurs due to the Poisson experiment, a statistical experiment and is usually measured in terms of success and failure. Poisson distribution is a special case of the binomial distribution. When the number of trials (n) increases suddenly even as μ = np, which denotes the expected value of several successful trials, the result remains constant. 

The Poisson Probability Mass Function is recognised for modelling the ‘number of times an event occurs in an interval of space and time’. 

The formula for Poisson Probability Mass function is - 

f (k; λ) = Pr (X=k) = λk e-λ/k!

Here ‘e’ = Euler’s number (2.71828…) whereas ‘k’ is defined as the number of occurrences of an event and ‘k!’ is the factorial of ‘k’. 

Poisson Distribution -  Mean and Variance 

While conducting a Poisson distribution experiment, the average number of successes in a given range is denoted as ‘λ’. Under Poisson distribution, the mean of a distribution is denoted by ‘λ’ and ‘e’ is constant. 

The Poisson probability, hence, would be -                                                  
P(x, λ ) =(e– λ λx)/x!

Under Poisson distribution, the mean is represented as E(X) = λ, and considering that mean and variance in it are equal, we get -                                      
  E(X) = V(X)

Here, V(X) will be the variance. 

This is an essential topic in the chapter called “Probability” of class 12. In past years’ board exams, we have noted that at least 1 question is asked. It covers 4 marks at most. Also, this topic is frequently noted in various national competitive exams too. 

1. Considering that the average number of soft drinks sold by a Kirana store is 2 bottles per day, what is the probability that exactly 3 soft drinks will be sold tomorrow?

Solution. We’re given the following details - 

μ = 2; 
x = 3; 
e (Euler’s  number) = 2.71828; 

Using Poisson Distribution formula -
P(x; μ) = (e-μ) (μx) / x!
P(3; 2) = (2.71828-2) (23) / 3!
P(3; 2) = (0.13534) (8) / 6
P(3; 2) = 0.180
Probability = 0.180 

2. Considering the probability of only 3 employees coming to the office today, what will be the probability that exactly 4 employees will come to the office tomorrow? 

Solution. As per Poisson distribution formula - 

Average rate of value(λ) = 3
Poisson random variable(x) = 4

P(X=x) = e-λλx/x!
P(X=4) = e-3 -34/4! 
P(X=4) = 0.16803135574154

3. A bike salesman sells an average of 3 bikes each week. Use Poisson’s formula to find the probability of him selling some bikes in a week. 

Solution. μ = 3; 

Some bikes imply = 1 or more bikes; so, 

P(X > 0) = 1 − P(x0)
P(X) = e−μ- μx/x! - P(x0 ) = e-3-30/0! = 4.9787x10-2 
= 1 - 4.9787x10-2 
 = 0.95021

Q: What is Poisson distribution?

Q: What is the ‘Poisson random variable’?

Q: What does ‘Euler’s number’ imply?

Q: Mention two conditions for a Poisson distribution experiment to take place.

Q: How do you classify results derived from a Poisson distribution experiment?