Higher Engineering Mathematics, Sixth Edition

The binomial and Poisson distributions 557

(Alternatively, the probability of having at least

1girlis:1−(theprobabilityofhavingnogirls),i.e.

1 − 0 .0625,giving0.9375,asobtainedpreviously.)

(b) The probability of having at least 1 girl and

1 boy is given by the sum of the probabilities of

having: 1 girl and 3 boys, 2 girls and 2 boys and 3

girls and 2 boys, i.e.

0. 2500 + 0. 3750 + 0. 2500 = 0. 8750

(Alternatively,this is also the probabilityof having

1 −(probabilityof having no girls+probabilityof

having no boys), i.e.

1 − 2 × 0. 0625 = 0. 8750 , as obtained previously.)

Problem 2. A dice is rolled 9 times. Find the

probabilities of having a 4 upwards (a) 3 times and

(b) less than 4 times.

Letpbe the probability of having a 4 upwards. Then

p= 1 /6,sincedicehavesixsides.

Letqbe the probability of not having a 4 upwards.

Thenq= 5 /6. The probabilities of having a 4 upwards

0 , 1 , 2 ,...,ntimes are given by the successive terms of

the expansion of(q+p)n, taken from left to right.From

the binomial expansion:

(q+p)^9 =q^9 + 9 q^8 p+ 36 q^7 p^2 + 84 q^6 p^3 +···

The probability of having a 4 upwards no times is

q^9 =( 5 / 6 )^9 = 0. 1938

The probability of having a 4 upwards once is

9 q^8 p= 9 ( 5 / 6 )^8 ( 1 / 6 )= 0. 3489

The probability of having a 4 upwards twice is

36 q^7 p^2 = 36 ( 5 / 6 )^7 ( 1 / 6 )^2 = 0. 2791

The probability of having a 4 upwards 3 times is

84 q^6 p^3 = 84 ( 5 / 6 )^6 ( 1 / 6 )^3 = 0. 1302

(a) The probability of having a 4 upwards 3 times is

0.1302

(b) The probability of having a 4 upwards less than 4

times is the sum of the probabilities of having a 4

upwards 0, 1, 2, and 3 times, i.e.

0. 1938 + 0. 3489 + 0. 2791 + 0. 1302 = 0. 9520

Industrial inspection

In industrial inspection, p is often taken as the

probability that a component is defective andqis the

probability that the component is satisfactory. In this

case, a binomial distribution may be defined as:

‘the probabilities that 0, 1, 2, 3,...,ncomponents

are defective in a sample of ncomponents, drawn

at random from a large batch of components, are

given by the successive terms of the expansion of

(q+p)n, taken from left to right’.

This definition is used in Problems 3 and 4.

Problem 3. A machine is producing a large

number of bolts automatically. In a box of these

bolts, 95% are within the allowable tolerance values

with respect to diameter, the remainder being

outside of the diameter tolerance values. Seven

bolts are drawn at random from the box. Determine

the probabilities that (a) two and (b) more than two

of the seven bolts are outside of the diameter

tolerance values.

Let pbe the probability that a bolt is outside of the

allowable tolerance values, i.e. is defective, and letqbe

the probabilitythat a bolt is within the tolerance values,

i.e. is satisfactory. Thenp=5%, i.e. 0.05per unit and

q=95%, i.e. 0.95per unit. The sample number is 7.

The probabilities of drawing 0, 1 , 2 ,...,ndefective

bolts are given by the successive terms of the expansion

of(q+p)n, taken from left to right. In this problem

(q+p)n=( 0. 95 + 0. 05 )^7

= 0. 957 + 7 × 0. 956 × 0. 05

+ 21 × 0. 955 × 0. 052 +···

Thus the probability of no defective bolts is

0. 957 = 0. 6983

The probability of 1 defective bolt is

7 × 0. 956 × 0. 05 = 0. 2573

The probability of 2 defective bolts is

21 × 0. 955 × 0. 052 = 0. 0406 ,and so on.

(a) The probability that two bolts are outside of the

diameter tolerance values is0.0406