Mathematics, 03.10.2019 04:30 jennifercastillo1233

This exercise illustrates that poor quality can affect schedules and costs. a manufacturing process has 70 customer orders to fill. each order requires one component part that is purchased from a supplier. however, typically, 2% of the components are identified as defective, and the components can be assumed to be independent. (a) if the manufacturer stocks 70 components, what is the probability that the 70 orders can be filled without reordering components? (b) if the manufacturer stocks 72 components, what is the probability that the 70 orders can be filled without reordering components? (c) if the manufacturer stocks 75 components, what is the probability that the 70 orders can be filled without reordering components?

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(a) 24.31%
(b) 24.31%
(c) 24.31%

Note: There is a common explanation for answers in questions (a) - (c) because the method used to answer these questions are same.

Explanation for (a) - (c):

For any positive integer n, let

a_n

= the event that the nth order has filled without reordering components
A_n

= the event that the n orders has filled without reordering components
B_n

= the event that the n components in manufacturer's stock has no defect

Since 2% of the components are defective, the probability that the component has defect is 2% or 0.02. Then, the probability that the component has no defect is 1 - 0.02 = 98%.

The order can only be filled without reordering component if and only if the component has no defect. Thus, the probability that the nth order has filled without reordering components is given by

P(a_n) = 0.98

Moreover, note that n orders are filled without reordering components if and only if for n consecutive times, the nth order has filled without reordering components.

In terms of equation,

$A_n = a_1 \cap a_2 \cap a_3 \cap ... \cap a_n$

Thus, the probability that n orders are filled without reordering components is given by

$P(A_n) = P(a_1 \cap a_2 \cap a_3 \cap ... \cap a_n)$

Since the components are independent, we can use the multiplication rule for independent events. Applying this rule, we obtain the following equation:

$P(a_1 \cap a_2 \cap a_3 \cap ... \cap a_n) = P(a_1)(P(a_2))(P(a_3))...(P(a_n))$

Since P(a_1) = P(a_2) = P(a_3) = ... = P(a_n) = 0.98

P(a_1) = P(a_2) = P(a_3) = ... = P(a_n) = 0.98

, therefore the probability that n orders are filled without reordering components is given by

$P(A_n) = P(a_1 \cap a_2 \cap a_3 \cap ... \cap a_n)
\newline P(A_n) = P(a_1)(P(a_2))(P(a_3))...(P(a_n))
\newline P(A_n) = (0.98)(0.98)(0.98)...(0.98)
\newline \boxed {P(A_n) = (0.98)^n}$

Explanation for (a):

Using what we defined earlier:

$A_{70}$ = Event that the n orders has filled without reordering components

$B_{70}$ = Event that the manufacturing company has 70 stocks

Then, the probability that the 70 orders can be filled without reordering components if the manufacturing company has 70 stocks is given by

$P(A_{70} | B_{70} ) = \frac {P(A_{70} \cap B_{70} )}{P(B_{70} )}$

Since the components are independent, $A_{70}$ and $B_{70}$ are also independent. So, by using the multiplication rule for independent events, we obtain the following:

$P(A_{70} \cap B_{70} ) = P(A_{70} )P(B_{70} )$

Thus,

$P(A_{70} | B_{70} ) = \frac {P(A_{70} \cap B_{70} )}{P(B_{70} )}
\newline P(A_{70} | B_{70} ) = \frac {P(A_{70})P(B_{70} )}{P(B_{70} )}
\newline \boxed {P(A_{70} | B_{70} ) = P(A_{70})}$

Moreover, using the computation we did earlier, the probability that the 70 orders can be filled without reordering components is given by

$P(A_{70}) = (0.98)^{70} \approx 0.2431$

Hence, the probability that the 70 orders can be filled without reordering components if the manufacturing company has 70 stocks is given by

$P(A_{70} | B_{70} ) = P(A_{70})
\newline \boxed{P(A_{70} | B_{70} ) \approx 0.2431 = 24.31 \% }$

Explanation for (b):

We can use the same method for (a). We define the following events:

$B_{72}$ = Event that the manufacturing company stocks 72 components without defect

Thus, using the same reasoning and computation in question (a), the probability that 70 orders are filled without reordering components given that the manufacturing company stocks 72 components without defect is given by:

$P(A_{70} | B_{72}) = \frac{P(A_{70} \cap B_{72})}{P(B_{72})} \newline P(A_{70} | B_{72}) = \frac{P(A_{70}) P(B_{72})}{P(B_{72})} \newline P(A_{70} | B_{72}) = P(A_{70}) \newline P(A_{70} | B_{72}) \approx 0.2431 \newline \boxed{P(A_{70} | B_{72}) \approx 24.31 \%}$

Explanation for (c):

Let

$B_{75}$ = event that the manufacturer stocks 75 components without defect.

Then, based on the reasoning used in question (a), the probability that 70 is filled without reordering components whenever event $B_{75}$ happens is given by

$P(A_{70} | B_{75}) = \frac{P(A_{70} \cap B_{75})}{P(B_{75})} \newline P(A_{70} | B_{75}) = \frac{P(A_{70}) P(B_{75})}{P(B_{75})} \newline P(A_{70} | B_{75}) = P(A_{70}) \newline P(A_{70} | B_{72}) \approx 0.2431 \newline \boxed{P(A_{70} | B_{75}) \approx 24.31 \%}$

Therefore, there is a 24.31% chance that 70 orders are filled without reordering components given that the manufacturing company stocks 75 components without defect.

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