Three plants manufacture hard drives and ship them to a warehouse for distributi

Question

Three plants manufacture hard drives and ship them to a warehouse for distribution. Plant I produces 55% of the warehouse’s inventory with a defect rate of 5%. Plant II produces 35% of the warehouse’s inventory with a defect rate of 8%. Plant III produces the remainder of the inventory with a 10% defect rate. a A warehouse inspectory selects one hard drive at random. What is the probability that it is a defective hard drive and from Plant II? b What is the probability that a randomly selected hard drive is defective? c Suppose a hard drive is defective. Which of the three plants is it most likely to have come from? Justify your answer using conditional probabilities.

Explanation / Answer

Let A shows the event that hard drive produce by Plant 1, B shows the event that hard drive produce by plant II and C shows the event that hard drive produce by Plant III. So

P(A) = 0.55, P(B) = 0.35, P(C) = 1 - 0.55 -0.35 = 0.10

Let D shows the event that hard drive is defective. So we have

P(D|A) = 0.05, P(D|B) = 0.08, P(D|C) = 0.10

a)

The probability that it is a defective hard drive and from Plant II is

P(D and B) = P(D|B)P(B) = 0.08 * 0.35 = 0.028

b)

The probability that a randomly selected hard drive is defective is

P(D) = P(D|A)P(A)+P(D|B)P(B) +P(D|C)P(C) = 0.05*0.55+0.08*0.35+0.10*0.10 = 0.0275+0.028+0.01=0.0655

c)

The probability that a defective hard drive come from plant I is

P(A|D) = [ P(D|A)P(A) ] / P(D) = 0.0275 / 0.0655 = 0.4198

The probability that a defective hard drive come from plant II is

P(B|D) = [ P(D|B)P(B) ] / P(D) = 0.028 / 0.0655 = 0.4275

The probability that a defective hard drive come from plant III is

P(C|D) = [ P(D|C)P(C) ] / P(D) = 0.01 / 0.0655 = 0.1527

That is it is most likley that it come from plant II.

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