Poisson Distribution

• Who invented “Poisson Distribution”?

A second important discrete probability distribution is the “Poisson Distribution”, named

after French mathematician S. Poisson who published its derivation in 1837.

• Characteristics of “Poisson Distribution”/ Why to use “Poisson Distribution”?

The characteristics of the Poisson distribution are as follows:

1. The occurrence of the events is independent. That is, the occurrence of an event in an
interval of space or time has no effect on the probability of a second occurrence of the
event in the same, or any other interval.

2. Theoretically, an infinite number of occurrences of the event must be possible in the

interval.

3. The probability of single occurrence of the event in a given interval is proportional to

the length of the interval.

4. In any infinitesimal (extremely small) portion of interval, the probability of two or more
occurrences of the event is negligible.

• Differences Between “Binomial Distribution” and “Poisson Distribution”:

“Poisson distribution” differs from the “Binomial distribution” in two important aspects:

1. “Binomial Distribution” is bi-parametric, i.e. it is featured by two parameters n and p,

whereas “Poisson Distribution” is uni-parametric (uniquely parametric or single
parametric), i.e. characterized by a single parameter m.

2. There are a fixed number of attempts (trials) in the Binomial Distribution. On the other
hand, an unlimited number of trials are there in a Poisson Distribution.

3. The success probability is constant in Binomial Distribution but in Poisson Distribution,

there are an extremely small number of success chances.

4. In a Binomial Distribution, there are only two possible outcomes, i.e. success or failure.
Conversely, there are an unlimited number of possible outcomes in the case of Poisson
Distribution.

5. In Binomial Distribution 𝑚𝑒𝑎𝑛 > 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒

while in Poisson Distribution 𝑚𝑒𝑎𝑛 = 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒.
6. Rather than consisting of discrete trials, the distribution operates continuously over some
given amount of time, distance, area, etc.

7. Rather than producing a sequence of successes and failures, the distribution produces
successes, which occur at random points in the specified time, distance, area. These
successes are commonly referred to as “occurrences”.

• Definition of “Poisson Distribution”:

Poisson distribution is a particular limiting form of the Binomial distribution when p (or
q) is very small and n (n=number of trials) is large.

Poisson distribution is given by

𝑒 −𝑚 𝑚 𝑥
𝑓(𝑥) = 𝑃(𝑥) = ; 𝑥 = 0, 1, 2, … … , 𝑛
𝑥!
where m is the mean of the distribution, x is the number of occurrences of the random
event, n is the number of trials and e is the constant whose value is 2.7183.
• Properies of “Poisson Distribution”:

The Poisson distribution satisfies the two essential properties, i.e. 𝑓(𝑥) ≥
0 𝑎𝑛𝑑 ∑ 𝑓(𝑥) = 1.

• Examples of “Poisson Distribution”:

The Poisson distribution has many applications in business and has been widely used in
management science and operations research. The following are some of the examples
which may be analyzed with the use of this distribution:

1. The demand for a product.

2. Typographical errors occurring on the pages of a book.
3. The occurrence of accident in a factory.
4. The arrival pattern in a departmental store.
5. The occurrence of flaws in a bolt in a factory.
6. The arrival of calls at a switch board.

• Formulas of “Poisson Distribution”:

1. The Mean: The mean of the Poisson distribution is given by

Mean=m=np

2. The Variance: The variance of the Poisson distribution is given by

Variance=m=np

3. The Standard Deviation: The standard deviation (S.D.) of the Poisson distribution is
given by
𝑆. 𝐷. = √𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = √𝑚 = √𝑛𝑝
 • Maths on “Poisson Distribution”:

Example 1: Assume that the probability of an individual coal miner being killed in a mine
1
accident during a year is 2400 . Use Poisson distribution to calculate the probability that in
a mine employing 200 miners, there will be at least one fatal accident in a year.

Solution: Given,
1
The probability of an individual coal miner being killed, 𝑝 = 2400. (probability p is very
low).
Number of miners working in the mine, 𝑛 = 200. (number of miners n is quite large).

Here, the probability p is very low and number of trials n is very large, that’s why, in this
case, we will apply Poisson Distribution.

𝑥 =Number of miners die/ Number of fatal accidents happen.

In this example, x=0, 1, 2, 3, …. …., n = 0, 1, 2, 3,… …, 200.
1 1
Mean, 𝑚 = 𝑛𝑝 = 200 × 2400 = 12.
𝑒 −𝑚 𝑚𝑥
By definition of Poisson distribution: 𝑃(𝑥) = ; 𝑥 = 0, 1, 2, … … , 𝑛.
𝑥!
According to question, 𝑃(𝑥 ≥ 1) =?.

𝑃(𝑥 ≥ 1) = 1 − 𝑃(𝑥 = 0)
1 1 0
𝑒 −12 (12)
=1−
0!
= 1 − 0.92 = 0.08 (𝑨𝒏𝒔)

Example 2: Suppose, one in 400 items is defective. If the items are packed in boxes of 100, what
is the probability that any given box of items will contain:

(i) No defectives;
(ii) Less than two defectives;
(iii) One or more defectives;
(iv) More than three defectives.

Solution: Here,

1
The probability of getting defective items, 𝑝 = 400. (probability p is very low).
Number of items, 𝑛 = 100. (number of items n is quite large).

Here, the probability p is very low and number of trials n is very large, that’s why, in this case, we
will apply Poisson Distribution.
 𝑥 =Number of defectives.
In this example, x=0, 1, 2, 3, …. …., n = 0, 1, 2, 3,… …, 100.
1
Mean, 𝑚 = 𝑛𝑝 = 100 × 400 = 0.25.

(i) Probability of no defectives=

𝑃(𝑥 = 0) = 𝑒 −𝑚 = 𝑒 −0.25 = 0.7788 (𝑨𝒏𝒔)

(ii) Probability of less than two defectives=

𝑃(𝑥 ≤ 1) = 𝑃(𝑥 = 0) + 𝑃(𝑥 = 1)
= 𝑒 −𝑚 + 𝑚𝑒 −𝑚 = 𝑒 −𝑚 (1 + 𝑚)
= 0.7788 (1 + 0.25) = 0.9735
(iii) Probability of one or more defectives
= 𝑃(𝑥 ≥ 1) = 1 − 𝑃(𝑥 = 0) = 1 − 𝑒 −𝑚 = 1 − 0.7788 = 0.2212

(iv) Probability of more than three defectives

= 𝑃(𝑥 ≥ 4) = 1 − [𝑃(𝑥 = 0) + 𝑃(𝑥 = 1) + 𝑃(𝑥 = 2) + 𝑃(𝑥 = 3)]
−𝑚 −𝑚
𝑚2 −𝑚 𝑚3 −𝑚
= 1 − [𝑒 + 𝑚𝑒 + 𝑒 + 𝑒 ]
2 6
𝑚2 𝑚3
= 1 − 𝑒 −𝑚 [1 + 𝑚 + + ]
2 6
= 1 − 0.7788 [1 + 0.25 + 0.03125 + 0.0026]
= 1 − 0.7788 [1.28385] = 1 − 0.99986 = 0.00014 (Ans)

Practice Problem 1: A factory produced blades in packets of 10. The probability of a blade to be
defective is 0.2%. Find the number of packets having two defective blades by applying Poisson
distribution.
Answer: 0.000196
Practice Problem 2: The probability rate that telephone calls arrive at a switchboard is 0.5 calls
per minute. Calculate the probability that two calls will arrive in a particular five-minute period
by using Poisson distribution.
Answer: 0.257
Practice Problem 3: The average number of calls received by a telephone operator during 5 P.M.
to 5.10 P.M. daily is 3. What is the probability that the operator will receive (i) no call (ii) one call
and (iii) at least two calls tomorrow during the same interval by applying Poisson distribution?
Answer: 0.0498, 0.1494, 0.8008