Question

This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fill. Each order requires one component part that is purchased from a supplier. However, typically, 3% of the components are identified as defective, and the components can be assumed to be independent.

Required:
a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?
b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?
c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

Answer 1

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

3% of the components are identified as defective

This means that
`p = 0.03`

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is
`P(X = 0)` when
`n = 90`. So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(90,0).(0.03)^(0).(0.97)^(90) = 0.0645`

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so
`P(X \leq 2)` when
`n = 102`

Then

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(102,0).(0.03)^(0).(0.97)^(102) = 0.0447`

`P(X = 1) = C_(102,0).(0.03)^(1).(0.97)^(101) = 0.1411`

`P(X = 2) = C_(102,2).(0.03)^(2).(0.97)^(100) = 0.2204`

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062`

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so
`P(X \leq 5)` when
`n = 105`

Then

`P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(105,0).(0.03)^(0).(0.97)^(105) = 0.0408`

`P(X = 1) = C_(105,0).(0.03)^(1).(0.97)^(104) = 0.1326`

`P(X = 2) = C_(105,2).(0.03)^(2).(0.97)^(103) = 0.2133`

`P(X = 3) = C_(105,3).(0.03)^(3).(0.97)^(102) = 0.2265`

`P(X = 4) = C_(105,4).(0.03)^(4).(0.97)^(101) = 0.1786`

`P(X = 5) = C_(105,5).(0.03)^(5).(0.97)^(100) = 0.1116`

`P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034`

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components