Higher Engineering Mathematics

554 STATISTICS AND PROBABILITY

Letqbe the probability of not having a 4 upwards.
Then q= 5 /6. The probabilities of having a 4
upwards 0, 1, 2,...,ntimes are given by the succes-
sive 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

is0.1302.

(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,pis often taken as the proba-

bility 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 defec-

tive in a sample ofncomponents, 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.

Letpbe the probability that a bolt is outside of the

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

qbe the probability that a bolt is within the toler-

ance values, i.e. is satisfactory. Thenp=5%, i.e.

0.05 per unit andq=95%, i.e. 0.95 per unit. The

sample number is 7.

The probabilities of drawing 0, 1, 2,...,ndefec-

tive 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.

(b) To determine the probability that more than

two bolts are defective, the sum of the proba-

bilities 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 com-

ponents 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. 0.08 per unit. Thus, the probability

of a component not being damaged,q,is1− 0 .08,

i.e. 0.92.