Higher Engineering Mathematics, Sixth Edition

560 Higher Engineering Mathematics

taken from left to right, i.e. by

e−λ,λe−λ,

λ^2 e−λ

2!

,...

Thus probability of no defective gearwheels is

e−λ=e−^2.^4 = 0. 0907

probability of 1 defective gearwheel is

λe−λ= 2 .4e−^2.^4 = 0. 2177

probability of 2 defective gearwheels is

λ^2 e−λ

2!

=

2. 42 e−^2.^4

2 × 1

= 0. 2613

(a) The probability of having 2 defective gearwheels

is0.2613

(b) The probability of having more than 2 defective

gearwheels is 1−(the sum of the probabilities 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 deter-

mine the theoretical probabilities whenp, the prob-

ability of an event happening, is known, butq,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 calculated, but it is not pos-

sible to quantify the number of goals which were not

scored. In this type of problem, a Poisson distribution

may be defined as follows:

‘the probabilitiesof an event occurring0, 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

breakdowns on each machine averages 0.06per

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 proba-

bilities 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 histogram

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 acci-

dent are given by the terms of expression

e−λ

(

1 +λ+

λ^2

2!

+

λ^3

3!

+···

)

,

taken from left to right.