Higher Engineering Mathematics

THE BINOMIAL AND POISSON DISTRIBUTIONS 557

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of having 0, 1, and 2 defective gearwheels), i.e.

1 −(0. 0907 + 0. 2177 + 0 .2613),

that is,0.4303

The principal use of a Poisson distribution is to
determine the theoretical probabilities whenp, the
probability of an event happening, is known, but
q, the probability of the event not happening is
unknown. For example, the average number of goals
scored per match by a football team can be calcu-
lated, but it is not possible to quantify the number
of goals which were not scored. In this type of
problem, a Poisson distribution may be defined as
follows:

‘the probabilities of an event occurring 0, 1, 2, 3, ...

times are given by the successive terms of the expression

e−λ

(

1 +λ+

λ^2

2!

+

λ^3

3!

+···

)

,

taken from left to right’

The symbolλis the value of the average occurrence
of the event.

Problem 7. A production department has 35

similar milling machines. The number of break-

downs on each machine averages 0.06 per week.

Determine the probabilities of having (a) one,

and (b) less than three machines breaking down

in any week.

Since the average occurrence of a breakdown is
known but the number of times when a machine did
not break down is unknown, a Poisson distribution
must be used.
The expectation of a breakdown for 35 machines
is 35× 0 .06, i.e. 2.1 breakdowns per week. The
probabilities of a breakdown occurring 0, 1, 2,...
times are given by the successive terms of the
expression

e−λ

(

1 +λ+

λ^2

2!

+

λ^3

3!

+···

)

,

taken from left to right.

Hence probability of no breakdowns

e−λ=e−^2.^1 = 0. 1225

probability of 1 breakdown is

λe−λ= 2 .1e−^2.^1 = 0. 2572

probability of 2 breakdowns is

λ^2 e−λ

2!

=

2. 12 e−^2.^1

2 × 1

= 0. 2700

(a) The probability of 1 breakdown per week is

0.2572.

(b) The probability of less than 3 breakdowns per

week is the sum of the probabilities of 0, 1, and

2 breakdowns per week,

i.e. 0. 1225 + 0. 2572 + 0 .2700, i.e. 0. 6497

Histogram of probabilities

The terms of a Poisson distribution may be repre-

sented pictorially by drawing a histogram, as shown

in Problem 8.

Problem 8. The probability of a person having

an accident in a certain period of time is 0.0003.

For a population of 7500 people, draw a his-

togram showing the probabilities of 0, 1, 2, 3, 4,

5 and 6 people having an accident in this period.

The probabilities of 0, 1, 2,...people having an

accident are given by the terms of expression

e−λ

(

1 +λ+

λ^2

2!

+

λ^3

3!

+···

)

,

taken from left to right.

The average occurrence of the event,λ,is

7500 × 0 .0003, i.e. 2.25.

The probability of no people having an accident is

e−λ=e−^2.^25 = 0. 1054

The probability of 1 person having an accident is

λe−λ= 2 .25e−^2.^25 = 0. 2371

The probability of 2 people having an accident is

λ^2 e−λ

2!

=

2. 252 e−^2.^25

2!

= 0. 2668

and so on, giving probabilities of 0.2001, 0.1126,

0.0506 and 0.0190 for 3, 4, 5 and 6 respectively hav-

ing an accident. The histogram for these probabilities

is shown in Fig. 57.2.