Higher Engineering Mathematics, Sixth Edition

558 Higher Engineering Mathematics

(b) To determine the probability that more than

two bolts are defective, the sum of the probabil-

ities of 3 bolts, 4 bolts, 5 bolts, 6 bolts and 7 bolts

being defective can be determined. An easier way

to find this sum is to find 1−(sum of 0 bolts, 1 bolt

and 2 bolts being defective), since the sum of all

the terms is unity. Thus, the probability of there

being more than two bolts outside of the tolerance

values is:

1 −( 0. 6983 + 0. 2573 + 0. 0406 ),i.e.0.0038

Problem 4. A package contains 50 similar

components and inspection shows that four

have been damaged during transit. If six

components are drawn at random from the contents

of the package determine the probabilities that in

this sample (a) one and (b) less than three are

damaged.

The probability of a component being damaged, p,

is 4 in 50, i.e. 0.08per unit. Thus, the probability of a

component not being damaged,q,is1− 0 .08, i.e. 0.92.

The probability of there being 0, 1 , 2 ,...,6 damaged

components is given by the successive terms of

(q+p)^6 , taken from left to right.

(q+p)^6 =q^6 + 6 q^5 p+ 15 q^4 p^2 + 20 q^3 p^3 +···

(a) The probability of one damaged component is

6 q^5 p= 6 × 0. 925 × 0. 08 =0.3164

(b) The probability of less than three damaged com-

ponents is given by the sum of the probabilitiesof

0, 1 and 2 damaged components.

q^6 + 6 q^5 p+ 15 q^4 p^2

= 0. 926 + 6 × 0. 925 × 0. 08

+ 15 × 0. 924 × 0. 082

= 0. 6064 + 0. 3164 + 0. 0688 =0.9916

Histogram of probabilities

The terms of a binomial distribution may be repre-

sented pictorially by drawing a histogram, as shown in

Problem 5.

Problem 5. The probability of a student

successfully completing a course of study in three

years is 0.45. Draw a histogram showing the

probabilities of 0, 1 , 2 ,...,10 students successfully

completing the course in three years.

Letpbe the probability of a student successfully com-

pleting a course of study in three years andqbe the

probability of not doing so. Then p= 0 .45 andq=

0 .55. The number of students,n, is 10.

The probabilitiesof 0, 1 , 2 ,...,10 students successfully

completing the course are given by the successive terms

of the expansion of(q+p)^10 , taken from left to right.

(q+p)^10 =q^10 + 10 q^9 p+ 45 q^8 p^2 + 120 q^7 p^3

+ 210 q^6 p^4 + 252 q^5 p^5 + 210 q^4 p^6

+ 120 q^3 p^7 + 45 q^2 p^8 + 10 qp^9 +p^10

Substitutingq= 0 .55 andp= 0 .45 in this expansion

gives the values of the successive terms as: 0.0025,

0.0207, 0.0763, 0.1665, 0.2384, 0.2340, 0.1596, 0.0746,

0.0229, 0.0042 and 0.0003. The histogram depicting

these probabilities is shown in Fig. 57.1.

0.24

0.22

0.20

0.18

0.16

0.14

0.12

Probability of successfully completing course

0.10

0.08

0.06

0.04

0.02

0

012345

Number of students

678910

Figure 57.1