Question

A large Computer Company A, subcontract the manufacturing of its circuit board to two Companies , 40% to Company B and 60% to Company C. Company B in turn subcontracts 70% of the orders it receives from Company A to Company D and the remaining 30% to Company E , both subsidiaries of Company B. When the boards are completed by companies D,E and C , they are shipped to Company A to be used in various computer models. It has been found that 1.5% , 1% and 0.5% of the boards from D,E and C respectively prove defection during the 90 day warranty period after a computer is first sold. What is the probability that a given board will be defective during the 90 day period?

Answer 1

Answer to Question #85442 - Math - Statistics and Probability

Question: A large Computer Company A, subcontract the manufacturing of its circuit board to two Companies , 40% to Company B and 60% to Company C. Company B in turn subcontracts 70% of the orders it receives from Company A to Company D and the remaining 30% to Company E , both subsidiaries of Company B. When the boards are completed by companies D,E and C , they are shipped to Company A to be used in various computer models. It has been found that 1.5% , 1% and 0.5% of the boards from D,E and C respectively prove defection during the 90 day warranty period after a computer is first sold. What is the probability that a given board will be defective during the 90 day period?

Solution: Let us define the following events:

- B: The event that the board came from company B

- C: The event that the board came from company C

- D: The event that the board came from company D

- E: The event that the board came from company E

It is clear that, $\mathbf{B=D\cup E}$ and the events $\mathbf{D}$ and $\mathbf{E}$ are disjoint. Also the events $\mathbf{B}$ and $\mathbf{C}$ are mutually exclusive and exhaustive.

Given a board, we want to find the probability that it will be defective during the 90 day period. Let $\mathbf{X}$ denote the event that the board will be defective in the warranty period.

$P(\mathbf{X})$

$=P(\mathbf{X\cap B})+P(\mathbf{X\cap C})$ [As, $\mathbf{B}$ and $\mathbf{C}$ are exclusive and exhaustive]

$=P(\mathbf{X\cap D})+P(\mathbf{X\cap E})+P(\mathbf{X\cap C})$

$=P(\mathbf{X|D})P(\mathbf{D})+P(\mathbf{X|E})P(\mathbf{E})+P(\mathbf{X|C})P(\mathbf{C})$

$=P(\mathbf{X|D})P(\mathbf{D|B})P(\mathbf{B})+P(\mathbf{X|E})P(\mathbf{E|B})P(\mathbf{B})+P(\mathbf{X|C})P(\mathbf{C})$

Observe that, all the values of the above expression has been given in the question. The values are,

$P(\mathbf{X|D})=\frac{1.5}{100}$

$P(\mathbf{X|E})=\frac{1}{100}$

$P(\mathbf{X|C})=\frac{0.5}{100}$

$P(\mathbf{D|B})=\frac{70}{100}$

$$P(\mathbf{E}|\mathbf{B}) = \frac{30}{100}$$ $$P(\mathbf{B}) = \frac{40}{100}$$ $$P(\mathbf{C}) = \frac{60}{100}$$

Substituting these values in the expression, we obtain,

$$\begin{aligned} & \frac{P(\mathbf{X})}{P(\mathbf{X})} \\ & = \frac{1.5}{100} \frac{70}{100} \frac{40}{100} + \frac{1}{100} \frac{30}{100} \frac{40}{100} + \frac{0.5}{100} \frac{60}{100} \\ & = \frac{4200 + 1200 + 3000}{100^3} \\ & = \frac{8400}{100^3} \\ & = 0.84\% \end{aligned}$$

Answer: Hence, the probability that a given board will be defective within 90 day period is $0.84\%$.

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