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Three plants manufacture hard drives and ship them to a warehouse for distribution. Plant I produces 54% of the warehouses inventory with a 4% defect rate. Plant II produces 35% of the warehouses inventory with an 8% defect rate. Plant III produces the remainder of the warehouses inventory with a 12% defect rate. (a) (5 points) A warehouse inspector selects one hard drive at random. What is the probability that a randomly selected hard drive is defectiv

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Answer:

A. 0.028

B. 0.0628

C. 0.45

Step-by-step explanation:

A. Calculation to determine the probability that it is adefective hard drive and from plant II

Using CONDITIONAL PROBABILITY formula

P(B∩A2) =P(A2)P(B|A2)

Let plug in the formula

P(B∩A2)= 0.35×0.08

P(B∩A2)= 0.028

Therefore the probability that it is adefective hard drive and from plant II will be 0.028

B. Calculation to determine the probability that a randomly selected hard drive is defective

Using LAW OF TOTAL PROBABILITY formula

P(B) =P(A1)P(B|A1) +P(A2)P(B|A2) +P(A3)P(B|A3)

Let plug in the formula

P(B)= 0.54×0.04 + 0.35×0.08 + 0.11×0.12

P(B)=0.0216+0.028+0.0132

P(B)= 0.0628

Therefore the probability that a randomly selected hard drive is defective is 0.0628

C. Calculation to determine the probability that it came from plant II

Using CONDITIONAL PROBABILITY formula two twice

P(A2|B) =P(A2∩B)/P(B)

Let plug in the formula

P(A2|B)=0.0280/0.0628

P(A2|B)= 0.45

Therefore the probability that it came from plant II is 0.45

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