Question

. A foreman for an injection-molding firm admits that on 23% of his shifts, he forgets to shut off the injection machine on his line. Failure to shut down at night causes the machine to overheat, increasing the probability that a defective molding will be produced during the early morning run from 5% to 15%. The plant manager randomly selects a molding from the early morning run and discovers it is defective. What is the probability that the foreman forgot to shut off the machine the previous night?

Answer 1

P(F | D) = 47.26%

There is a 47.26% probability that the foreman forgot to shut off the machine the previous night.

Explanation:

A foreman for an injection-molding firm admits that on 23% of his shifts, he forgets to shut off the injection machine on his line.

Let F denote the event that foreman forgets to shut off the machine.

Failure to shut down at night causes the machine to overheat, increasing the probability that a defective molding will be produced during the early morning run from 5% to 15%.

Let D denote the event that the mold is defective.

If the foreman forgets to shut off the machine then 15% molds get defective.

P(F and D) = 0.23×0.15

P(F and D) = 0.0345

If the foreman doesn't forget to shut off the machine then 5% molds get defective.

P(F' and D) = (1 - 0.23)×0.05

P(F' and D) = 0.77×0.05

P(F' and D) = 0.0385

The probability that the mold is defective is

P(D) = P(F and D) + P(F' and D)

P(D) = 0.0345 + 0.0385

P(D) = 0.073

The probability that the foreman forgot to shut off the machine the previous night is given by

∵ P(B | A) = P(A and B)/P(A)

For the given case,

P(F | D) = P(F and D)/P(D)

Where

P(F and D) = 0.0345

P(D) = 0.073

So,

P(F | D) = 0.0345/0.073

P(F | D) = 0.4726

P(F | D) = 47.26%