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

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

22 and 22

step-by-step explanation:

10 plus 12 equals 22.

12 plus 10 also equals 22 because regardless of what order they're in, they're gonna equal the same thing.

rounding to the nearest thousandth would be 452,000