The document discusses the Poisson distribution and its properties. It provides examples of how to calculate probabilities of events occurring using the Poisson distribution. Specifically, it discusses how the Poisson distribution can be used to model random, independent events that occur at a constant average rate over time or space. It also describes how the Poisson distribution can approximate the binomial distribution for large sample sizes and small probabilities.

1. The Poisson Distribution

2. If we are interested in the number of occurrences
that take place in an interval (time/volume)
Then we use
the Poisson Distribution

3. Situations to be modeled by Poisson
Distribution are,
▪ Events occur singly
▪ The probability that an event occurs is
the same for each interval/ at the
constant rate.
▪ Events occur independently

4. The variable X is the number of occurrences in
the given interval and we write
X ~ Po (λ) where λ is the parameter
and λ > 0.
When the Poisson Distribution is appropriate, the
probability of exactly x occurs is given by the
formula
P( X = x) = e- λ λx
x!

5. Example:
An average of 3 cars arrive at a highway tollgate every minute. If
this rate is approximately Poisson Distribution what is the
probability that exactly,
(a) 5 cars will arrive in one minute
(a) 7 cars will arrive in 5 minutes
Solution:
If X is the number of cars that arrive at the tollgate, then
X ~ Po (λ) where λ = 3.
(a) P( X = 5) = = 0.1008
(b) λ = 15, P( X = 7) = = 0.0104

6. Example :
Emergency calls to an ambulance service
are received at random times, at an average of
2 per hour. Calculate the probability that, in a
randomly chosen one hour period,
(a) no emergency calls are received
(b) exactly one call is received in the first
half hour and exactly one call is received
in the second half.

7. Solution:
a) If X is the number of emergency calls received, then
X ~ Po (λ) where λ = 2
P( X = 0) = e-220 = 0.1353
0!
b) Suppose Y1 : the number of emergency call received in the
first half and
Y2 : ……… the second half and λ = 1.
P(Y1 = 1) = e-111
1!
and
P(Y2 = 1)= e-111
1!
P(Y1 = 1) x P(Y2 = 1) =0.1353

8. Mean and Variance of Poisson Distribution.
If X is a random variable following the Poisson distribution
X ~ P(λ),
then,
the mean of X is E(X) = λ
and variance Var(X) = λ.

9. Example :
If X ~ Po (1.8),
the mean, E(X) = 1.8 and
the variance, Var(X) = 1.8.

10. Poisson Distribution as Approximation to
Binomial Distribution.
Under certain conditions the Poisson distribution can be used as an
approximation to the binomial distribution.
The important factors are the values of n and p.
If n is large (n > 50) and p is small (p < 0.1),
Then
X ~ B(n , p) can be approximate to
X ~ Po (λ), where λ = np.

11. Example :
There are 500 eggs in a box. At an average,
0.06% are broken before it arrives at a
supermarket.
Find the probability in a box containing 500
eggs,
(a) exactly 3 eggs are broken
(a) less than 2 eggs are broken
(c) at least 3 eggs are broken

12. Solution:
X is a random variable of the number of eggs broken.
n =500 (n > 50) and
p= 0.0006 (p < 0.01),
we can use the approximation to Poisson.
X ~ B(500 , 0.006) → X ~ Po ( λ )
where λ = np = 500 x 0.006 = 3, so that
X ~ Po (3).
(a) P( X = 3) = = 0.2240
(a) P( X < 2) = 1 – P(X ≥ 2)
= 1 – 0.8009
= 0.1991
(c) P(X ≥ 3) = 0.5768

13. Example :
The probability that an adult suffers an allergic reaction to a particular
inoculation is 0.0018.
If 5000 adults are given the inoculation, estimate the probability that ,
(a) at most 10 suffer an allergic reaction.
(b) From 5 to 15 inclusive suffer an allergic reaction.

14. Solution :
Y :number of adults suffers, Y~ B(5000, 0.0018).
n = 5000 (n > 50) p = 0.0018 (p < 0.01),
the approximation to Poisson can be used.
Y~ B(5000 , 0.0018) → X ~ Po (λ ) where λ = np = 5000 x 0.0018 = 9,
so that
X ~ Po (9)
(a) P( X ≤ 10) = 1 – P(X > 11)
= 1 – 0.2940
= 0.7060
(b) P( 5 ≤ X ≤ 15) = P(X ≥ 5) – P(X ≥ 16)
= 0.9450 – 0. 0220
= 0.9230

15. Example :
In a town, one out of 100 residents has group A blood.
(a)400 blood donors are taken at random.
By using the suitable approximation, find the probability that
more than 4 donors are having group A blood.
(b)Find the minimum number of donors that must be taken at
random, so that the probability of at least one donor having
group A blood is 0.8.

16. Solution :
(a)Suppose that X is the number of blood A resident, then
X ~ B(400,0.01).
n = 400 (large) p = 0.01 (small),
Poisson approximation.
λ = np = 400 x 0.01 = 4, X ~ Po (4)
P( X ≥ 5) = 0.3712
(b) P(X ≥ 1) = 1 – P ( X < 1)
= 1 – P ( X = 0 )
0.8 = 1 – e-0.01n
e-0.01n = 0.2
n = n ≈ 161

17. Example
Fanfold paper for computer printers is made by putting
perforations every 30 cm in a continuous roll of paper.
A box of fanfold paper contains 2000 sheets. State the length of
the continuous roll from which the box of paper is produced.
The manufacturers claim that faults occur at random and at
average rate 1 per 240 metres of paper.
State an appropriate distribution for the number of faults per box
of paper. Find the probability that it has more than 4 faults.
Two copies of a report which runs to 100 sheets per copy are
printed on this sort of paper.
Find the probability that there are no faults in either copy of the
report and also the probability that just one copy is faulty .

18. Solution
No of sheets = 2000
Length = 2000 X 30
= 60 000cm
= 600 metres
Faults : 1 per 240
No of faults per box = 600/240 = 2.5
So suitable distribution is X ~ Po (2.5)
P ( X=0) = ℮ -2.5 = 0.0821 ( 3 sf)
P ( X > 4) = P( X ≥ 5 )
= 0.1088
= 0.109 ( 3 s f )

19. 100 sheets = 100X 30 = 3000 cm = 30 m
So no. of faults = 30/240 = 1/8 = 0.125
→ Y ~ Po ( 0.125 )
P ( Y = 0 ) = ℮ -0.125 = 0.882496902=0.88250
→ P ( Both reports have no faults ) = ℮ -0.125 X ℮-0.125
= 0.7788
= 0.779 ( 3 sf)
P ( just one faulty ) = 2( 0.8825 )( 1 – 0.8825 )
= 0.20739
= 0.207 ( 3 s f )
X=0
X ≠0
Y=0
Y≠0
Y=0
Y ≠0

20. Example
Customers enter an antique shop independently of one another and at random
intervals of time at an average rate of four per hour throughout the five days
of a week on which the shop is open.
The owner has a coffee-break of fifteen minutes each morning; if one or more
customers arrive during this period then his coffee goes cold, otherwise he
drinks it while it is hot.
Let X be the random variables denoting the number of customers arriving
during a Monday coffee – break , and
let Y be the random variables denoting the number of days during a week
on which the owner’s coffee goes cold.
Assuming that X has a Poisson distribution, determine ( correct to 3 significant
figures)
a) P( X=0)
b) P( X≥ 2)
c) E( Y )
d) P( Y = 2 )

21. Solution
15 minutes break
4 per hour 🡪 1 per 15 minutes
X ~ P ( 1)
P ( X = 0) = ℮-1
= 0.36787
= 0.368 ( 3 s f)
P ( X ≥ 2 ) = 1 - P ( X ≤ 1)
= 1 – [ P ( X = 0 ) + P ( X = 1 ) ]
= 1 – [℮ -1 + ℮ -1 ]
= 0.264 ( 3 s f)
P ( coffee goes cold) = P ( X≥ 1)
= 1 – P ( X = 0)
= 0.63212
Y ~ Bin ( 5, 0.63212)
E ( Y )= 3.1606= 3.16 ( 3 s f)
d) P ( Y = 2 ) = 5 C 2 ( 0.632122 ) ( 0.367872 )
= 0.19892
= 0.199 ( 3 s f)