Question

A manufacturing process has 45 customer orders to fill. Each order requires one component part that is purchased from a supplier. However, typically 6% of the components are identified as defective, and the components can be assumed to be independent. (a) If the manufacturer stocks 47 components, what is the probability that the 45 orders can be filled without reordering components?

Answer 1

Answer:

Given that:

A manufacturing process has 45 customer orders to fill. Each order requires one component part that is purchased from a supplier. However, typically, 6% of the components are identified as defective, and the components can be assumed to be independent.

If the manufacturer stocks 47 components, what is the probability that the 45 orders can be filled without reordering components?

Let X be the number of good components in a sample of 47 components. Probability that selected component is non-defective, p = 1-6 %. 0.94

Here,

$X \sim Binom(n = 47, p = 0.94)$

Then, the PMF of X can be written as

$P(X= x) = \binom{n}{x}p^{x}(1-p)^{n-x}$

$P(X= x) = \binom{47}{x}(0.94)^{x}(0.06)^{47-x}$ x = 0,1,2...47

a) Calculate the probability that the 45 orders can be filled without reordering components

$P(X= 45) = \binom{47}{45}(0.94)^{45}(0.06)^{47-45}$

$P(X= 45) = \binom{47}{45}(0.94)^{45}(0.06)^2$

$P(X= 45) = 1081* 0.06176* 0.0036$

$P(X= 45) = 0.2403$