Poisson Random Variables

The Poisson random variables is also called as Poisson law of large numbers. Poisson variable is named after the French Mathematician Simeon Denis Poisson (1781 − 1840) who discovered it. Poisson distribution is also a discrete distribution. Poisson distribution is a some limiting case of Binomial distribution under the following conditions.

(i) n the number of trials is indefinitely large ie., n → ∞.

(ii) p is represented by constant probability of success in each trial is very small ie., p → 0.

(iii) np = λ is finite where λ is a positive real number. When this event rarely occurs, the distribution of such an event may be assumed to follow a Poisson distribution. The poisson random variables examples and practice problems are given below.

Example problem for poisson random variables:

Example problem:

If the number of incoming autos per minute at a auto terminus is a random variable having a Poisson distribution with λ=0.9, find the probability that there will be

(i) Exactly 9 incoming autos during a period of 5 minutes

(ii) Fewer than 10 incoming autos during a period of 8 minutes.

Solution:

λ for number of incoming autos per minute = 0.9

λ for number of incoming autos per 5 minutes = 0.9 × 5 = 4.5

P exactly 9 incoming autos during 5 minutes = (e−λ λ9)/ 9!

i.e., P(X = 9) =[ e−4.5 × (4.5)9 ] / 9!

Fewer than 10 incoming autos during a period of 8 minutes = P(X <10 p="">
Here λ = 0.9 × 8 = 7.2

Required probability = Σ  e−7.2 × (7.2)x] /x!  Limits x=0 to 9

Practice problems for Poisson random variables:

Practice problem 1:

The number of accidents in a year involving taxi drivers in a city follows a Poisson distribution with mean equal to 3. Out of 1000 taxi drivers find the approximate number of drivers with

(i) no accident in a year

(ii) more than 3 accidents in a year [e−3 = 0.0498].

Answer: (i) approximately 50 drivers

(ii) approximately 353 drivers

Practice problem 2:

Alpha particles are emitted by a radio active source at an average rate of 5 in a 20 minutes interval. Using Poisson distribution find the probability that there will be

(i) 2 emission

(ii) at least 2 emission in a particular 20 minutes interval. [e−5 = 0.0067].

Answer: (i) 0.0838 (ii) 0.9598

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