A Short Note on Poisson Probability Distribution

The Poisson distribution is a discrete probability characteristic meaning the variable can best take unique values in a given listing of numbers, probably infinite. Poisson distribution measures how usually an event is possible to occur within the “x” time frame.

A Poisson random variable “x” defines the number of successes within the test. This distribution takes place when there are activities that do not occur because of the consequences of a precise quantity of outcomes. Poisson distribution is used below certain conditions. They’re:

• The range of trials “n” tends to infinity
• The probability of success “p” tends to zero
• Np = 1 is finite

Poisson Probability Distribution Formula-

The formula for the Poisson probability distribution function is defined as follows -: 

F(x) =(e^– λ ^x)/x!

Where,

E is the base of the logarithm

X is a Poisson random variable

Λ is an average rate of value

Characteristics of a Poisson Probability Distribution-

Below are a few characteristics of the Poisson probability distribution –

• The test includes counting the number of events to be able to occur for the duration of a specific interval of time or in a specific distance, area, or extent.
• The probability that an event occurs in a given time, distance, location, or extent is identical.
• Every event is impartial to all different events. For an instance, the range of folks that arrive inside the first hour is unbiased of the number who arrive in some other hour.

Examples of Poisson Probability Distribution-

Example 1 -:  A random variable Y has a Poisson distribution with parameter λ such that P ( Y= 1) = (0.2) P (Y = 2). find P (Y = zero).

Solution -: For the Poisson distribution, the probability function is defined as: P (Y =x) = (e– λ λx)/x!, where λ is a parameter.

Given that, P (Y = 1) = (0.2) P (Y = 2)

(e^– λ .λ^1)/1! = (0.2)(e^– λ λ2)/2!

⇒λ = λ2/ 10

⇒λ = 10

Now, putting λ = 10, in the formula, we get:

P (Y=0 ) = (e– λ λ0)/0!

P (Y=0) = e-10 = 0.0000454

Thus, P (Y= 0) = 0.0000454

Example 2 -: phone calls arrive at a trade in line with the Poisson technique at a price λ= 2/min. Calculate the probability that exactly two calls may be acquired in the course of each of the primary 5 minutes of the hour.

Solution -: Let us assume that “N” be the number of calls received during a 1 minute period.

So ,

P(N= 2) = (e-2. 22)/2!

P(N=2) = 2e-2.

Now, “M” be the number of minutes amongst 5 minutes considered, at some stage in which precisely 2 calls could be received. Thus “M” follows a binomial distribution with parameters n=5 and p= 2e^-2.

P(M=5) = 32 x e^-10

P(M =5) = 0.00145, where “e” is a constant, which is  equal to 2.718. 

Poisson distribution and Binomial Distribution- 

Whether to use binomial distribution or Poisson distribution, it is important to understand the rule so that it becomes easy to understand which one to apply. The average probability of the event is there per unit, and to find the probability of a certain number of events that are happening in a specific period of time, then use the Poisson Distribution. If the exact probability is already given in the question and it is asked to find the probability of an event happening a specific number of times out of x given times, then use the binomial distribution to solve the question. 

Uses of Poisson Distribution-

Poisson distribution is used in business. It is used by businessmen to forecast sales. It tells the number of customers or sales on specific days or seasons of the year. It is not good to overstock the goods because it might lead to a loss in the business. On the other hand, if the stocks would be less, then also there is a possibility of loss in the business because there are not enough stocks to sell as per the demand in the market. By using the Poisson distribution, a businessman will be able to calculate when the demand will be higher so that they can purchase the goods. With the Poisson distribution, companies can estimate the supply to demand so that their business will be successful and will help in preventing the loss. 

Conclusion -: 

We will say that the Poisson Distribution is useful in uncommon occasions where the probability of fulfillment (p) could be very small and the possibility of failure (q) is very large and the value of n is very big. We have also discussed the characteristics of the Poisson distribution and explained the concept with examples to understand it in a better way.