1
CHAPTER:
THE POISSON DISTRIBUTION
Contents
0
Introduction
1
Poisson Distribution
2
Expectation and Variance
3
Poisson Distribution as an Approximation to Binomial Distribution
4
The Recurrence Formula
5
The Most Probable Value of X
6
Finding a Theoretical Distribution
7
Distribution of Two Independent Poisson Variables
8
Miscellaneous Examples
 0 Introduction
If an event is
a) randomly and uniformly scattered in time or space,
b) has mean number of occurrence, , in a given fixed interval of time or space,
c) the mean, in any interval is proportional to the length of the interval, and
d) X is the random variable ‘the no. of occurrences in the given interval’,
then X follows a Poisson distribution.
Examples of events which might follow a Poisson Distribution:
 the number of car accidents along Whitley Road in one year
 the number of telephone calls made to a switchboard in a given minute
 the number of bacteria in 1 ml of a liquid - - - - - - etc
 1 Poisson Distribution
A discrete random variable X which follows a Poisson distribution has probability density function of the form
x
P(X = x) = e-
for x = 0, 1, 2, 3, ... to infinity, where  > 0 is a positive constant.
x!
We write
X  Po( )
Note:  is the parameter of the distribution.
Example 1.1
The mean number of bacteria per millilitre of a liquid is known to be 4. Assuming that the number of bacteria
follows a Poisson distribution, find the probability that, in 1 ml of liquid, there will be
a) no bacteria
b) 4 bacteria
c) less than 3 bacteria.
[ 0.0183, 0.195, 0.238]
Solution
2
 2 Expectation and Variance
If
X ~ Po( ) ,
then E(X) =
and Var(X) = 
Since a Poisson distribution is uniformly scattered, the mean, in any interval, is proportional to the length of the
interval. Hence, E(X) =  can be modified proportionately to suit the given interval.
Example 2.1 (NJC 00/2/8 part, modified)
Independently for each page of a Mathematics textbook, the number of typing errors, Y, has a Poisson
distribution. Given that P(Y = 3 ) = 5 P(Y = 4), show that E(Y) is 0.8.
Solution
Example 2.2
The mean number of bacteria per millilitre of a liquid is known to be 4. Assuming that the number of bacteria
follows a Poisson distribution, find the probability that
a) in 3ml of liquid there will be less than 2 bacteria,
1
b) in ml of liquid, there will be more than 2 bacteria.
[ 7.99 x 10-5, 0.323]
2
Solution
3
Example 2.3
The number of goals scored in a match by a CJC team follows a Poisson Distribution with mean .
If the probability that the team scores no goals in a match is 0.301, find
a) the value of 
b) the probability that the team scores less that 3 goals in a match
c) the probability that the team scores less than 3 goals in 2 matches. [ 1.2, 0.879, 0.570]
Solution
 3 Poisson Distribution as an Aproximation to Binomial Distribution
Since the Poisson distribution arises from the Binomial distribution as a limiting case as n , p  0, in such a
way that np =  (a constant), it gives a good approximation to the Binomial distribution for large n and small p.
So, if a random variable X has a binomial distribution given by X ~ Bin(n,p) and n is large and p is small,
then X is approximately a Poisson random variable with mean  = np i.e.
X ~ Po(np)
Generally, we use the approximation when either n > 50 and p < 0.1 or n > 50 and np < 5.
Trivia:
Poisson distribution was derived from the Binomial distribution by Poisson, a 19th century French
mathematician Simeon-Denis Poisson (1781-1840). He was also a lecturer, a professor and examiner who made
contributions to mathematical physics eg. the Poisson ratio in elasticity and Poisson’s constant in electricity.
Example 3.1
A factory packs eggs in boxes of 500. On average, the probability that an egg is found to be broken when the
eggs are unpacked is 0.002. Find the probability that a box contains 2 broken eggs. [ 0.184]
Solution
4
Example 3.2
On average one in 200 cars breaks down on a certain stretch of road per day. Find the probability that on a
certain day
a) none of a sample of 250 cars breaks down,
b) more than 2 of a sample of 300 cars break down.
[ 0.29, 0.19]
Solution
Example 3.3
Eggs are packed in boxes of 500. On average, 0.8% of the eggs unpacked are broken.
a) Find the probability that in a box of 500 eggs,
i) exactly 3 will be broken
ii) less than 2 will be broken.
b) A supermarket unpacks 100 boxes of eggs. What is the probability that there will be exactly 4 boxes
containing no broken eggs? [ 0.195, 0.0916, 0.075]
Solution
5
Example 3.4
An aircraft has 116 seats. The airline has found, from long experience, that on average 2.5% of people with
tickets for a particular flight do not arrive for that flight. If the airline sells 120 tickets for a particular flight,
determine, using a suitable approximation, the probability that more than 116 people arrive for that flight.
Determine also the probability that there are empty seats in the flight. [ 0.647, 0.185]
Solution
Example 3.5
Samples each of 15 articles, are taken at random from a large consignment in which 20% of the articles are
defective. Find the probability that a random sample of 15 articles will contain
a) no defective articles and
b) at least 3 defective articles.
If 50 samples of 15 articles are to be examined, calculate the probability that less than 3 samples have no
defective articles. [ 0.0352, 0.602, 0.742]
Solution
6
 4 The Recurrence Formula
x
x+1
If X ~ Po(), then P(X = x) = e-  . So, P(X = x + 1) = e- 
.
x!
(x + 1)!
x+1
x
P(X = x + 1)
Dividing, we get
= e- 
 e- 
P(X = x)
(x + 1)!
x!
x+1

x!
= e-
.
(x + 1)! e- x
= 
x+1
or,
P(X = x + 1)
=  P(X = x)
x+1
This is the recurrence formula for the Poisson distribution.
Example 4.1
If X ~ Po(2.3), use the recurrence formula to find P(X  5). [ 0.0837]
Solution
for x = 0, 1, 2, ...
for x = 0, 1, 2, ...
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 5 The Most Probable Value of X
The value of X that is most likely to occur is the one with the highest probability.
There are 3 possible methods:
1) By observation –
When E(X) = k, where k is an integer, then the most probable values of X are k - 1 and k.
When E(X) = k, where k is a decimal, then the most probable value of X is [ k ], the integer value just less
than the decimal number of k. (‘round down’)
For example, when  = 4.7, then the value of X that is most likely to occur is 4.
But this method is not foolproof. It just serves as a guide to the correct value(s).
2) By testing the mean –
To find the most likely value of X, take the expectation of X and test the probabilities around that value.
This is possible because the mode is always somewhere around the mean for a Poisson distribution.
3) By using the recurrence formula
Example 5.1
If X  Po(3.5), find the most likely value of X. [ 3]
Solution
8
Example 5.2
If X ~ Po(3), find the most likely value of X. [2 and 3]
Solution
 6 Finding a Theoretical Distribution
It is sometimes useful to compare experimental results with a theoretical distribution.
Example 6.1
The following table shows the number of telephone calls received over a period of 150 days:
No. of calls
0
1
2
3
4
No. of days
51
54
36
6
3
a) Find the average number of calls per day [1.04]
b) Calculate the frequencies of the comparable Poisson distribution.
Solution
9
 7 Distribution of Two Independent Poisson Variables
If X and Y are two independent Poisson variables such that
then
X + Y  Po (m + n).
X  Po(m),
Y  Po(n),
Note: X - Y is not a Poisson distribution.
Example 7.1
Telephone calls reach a secretary independently and at random, internal ones at a mean rate of 2 in any 5
minutes period, and external ones at a mean rate of 1 in any 5 minutes period. Calculate the probability that
there will be more than 2 calls in any period of 2 minutes. [0.121]
Solution
Example 7.2
The centre pages of the Weekly Sentinel consists of 1 page of film and theatre reviews and 1 page of classified
advertisements. The number of misprints in the review has a Poisson distribution with mean 2.3 and the number
of misprints in the classified section has a Poisson distribution with mean 1.7.
a) Find the probability that, on the centre pages, there will be
i)
no misprints
ii)
more than 5 misprints.
b) Find the smallest integer n such that the probability that there are more than n misprints on the centre pages
is less than 0.1. [ 0.018, 0.215, 7]
Solution
10
Example 7.3
A building has 2 independent automatic telephone exchanges A and B. The number X of wrong connection for
A in any one day is a Poisson variable with parameter , and the number Y of wrong connections for B in any
one day is also a Poisson variable with parameter 2.
a) Find, to 3 decimal places, the largest value of  for which the probability of one or more wrong connections
1
for A in any day is at most .
6
b) It is given that  = 0.5. Calculate in any particular day,
i) the probability that there will be exactly three wrong connections in the building,
ii) the probability that there will be at most 3 wrong connections in the building given that X  2,
iii) the least integer r such that the probability of having at most r wrong connections in the building is
greater than 0.95.
[ 0.1823, 0.1255, 0.6699, 4]
Solution
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 8 Miscellaneous Examples
Example 8.1
On average a Coast guard station receives one distress call every 2 days.
A 'bad' week is a week in which 5 or more distress calls are received.
Show that the probability that a week is a bad week is 0.275 (correct to 3 sig. fig.).
Find the probability that, in 8 randomly chosen weeks, at least 2 are bad weeks. [ 0.692]
Solution
Example 8.2 [N84/P2/11]
An administration centre has 3 independent telephone lines A, B and C. During the period 10 00 to 10 15 hours
on a working day, the number of phone calls coming in on lines A, B and C are X1, X2 and X3 respectively and
each of these independent random variable has a Poisson distribution. Also it is known that
E(X1) = 1.2, E(X2) = 1.5 and that on 1 in 200 working days, there are no phone calls made to the centre between
10 00 and 10 15 hours. Find, correct to 3 significant figures,
a) the value of E(X3),
b) the probability that, on any given working day, there will be exactly 1 incoming call to the centre between
10 00 and 10 15 hours.
c) the probability that, on not more than 2 out of 100 working days, there will be exactly 1 incoming call to the
centre between 10 00 and 10 15 hours. [ 2.60, 0.0265, 0.506]
Solution
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Example 8.3
During each working day in a certain factory, a number of accidents occur independently according to a Poisson
distribution with mean 0.5. Calculate the probability that
a) during any one day, there are 2 or more accidents,
b) during two consecutive days, there are exactly 3 accidents altogether.
Out of 50 consecutive 5-day weeks, how many weeks would you expect to be accident free?
[0.0902, 0.0613, 4]
Solution
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SUMMARY
If an event is
a) randomly and uniformly scattered in time or space,
b) has mean number of occurrence, , in a given fixed interval of time or space,
c) the mean, in any interval is proportional to the length of the interval, and
d) X is the random variable ‘the no. of occurrences in the given interval’,
then X follows a Poisson distribution.
Notation:
X  Po( )
Probability distribution function :
P(X = x)
Expectation:
Variance:
E(X)
Var(X)
Recurrence formula:
= e-
x
x!
= 
= 
P(X = x+1) =  P(X=x)
x+1
for x = 0, 1, 2, 3, ... to infinity
for x = 0, 1, 2, ...
The Poisson distribution is used as an approximation to the Binomial distribution
when n > 50 and p < 0.1 or n > 50 and np < 5.
The Most Probable Value of X
The value of X that is most likely to occur is the one with the highest probability.
There are 3 possible methods:
1) By observation –
When E(X) = k, where k is an integer, then the most probable values of X are k - 1 and k.
When E(X) = k, where k is a decimal, then the most probable value of X is [ k ], the integer value just less
than the decimal number of k. (‘round down’)
For example, when  = 4.7, then the value of X that is most likely to occur is 4.
But this method is not foolproof. It just serves as a guide to the correct value(s).
2) By testing the mean –
To find the most likely value of X, take the expectation of X and test the probabilities around that value.
This is possible because the mode is always somewhere around the mean for a possion distribution.
3) By using the recurrence formula
If X and Y are two independent Poisson variables such that X  Po(m), Y  Po(n),
then X + Y  Po(m + n).
The Handshake Problem
Mr and Mrs Lee recently attended a party at which there were three other couples. Various handshakes
took place. No one shook hands with his or her own spouse, no one shook hands with the same person twice,
and of course, no one shook his or her own hand.
After all the handshaking was finished, Mr Lee asked each person, including his wife, how many hands
he or she had shaken. To his surprise, each gave a different answer. How many hands did Mrs Lee shake?
Puzzle taken from lectures by Assoc Prof Tan Eng Chye, NUS Maths Dept
“Life is good for only 2 things; to do maths and to teach it.” Simeon-Denis Poisson.