Question

A foreman for an injection-molding firm admits that on 9% of his shifts, he forgets to shut off the injection machine on his line. This causes the machine to overheat, increasing the probability that a defective molding will be produced during the early morning run from 1% to 19%.

Let F = forgets to shut off the injection machine and D = defective.

Use a probability tree to answer the following questions:

(Round all answers to four decimals)

a) What’s the probability a molding from the early morning run is defective?

b) 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

Introduction:

From the given info, P (F) = 0.09.

Therefore, P (F^C) = 1 – P (F) = 1 – 0.09 = 0.91.

When the foreman does not forget to shut the machine, probability of getting a defective molding is 1%. That is, P (D | F^C) = 0.01.

When the foreman forgets to shut the machine, probability of getting a defective molding becomes 19%. That is, P (D | F) = 0.19.

Calculation:

a.

The probability that a molding from the early morning run is defective, is:

P (D) = P (D ∩ F) + P (D ∩ F^C).

Conditional probability:

For two events A and B, the probability of the occurrence of A given that event B has already occurred, is: P (A | B) = P (A ∩ B) / P (B). From this conditional probability, it is evident that: P (A ∩ B) = P (A | B) P (B).

Thus, P (D ∩ F) = P (D | F) P (F) = (0.19) (0.09) = 0.0171.

Again, P (D ∩ F^C) = P (D | F^C) P (F^C) = (0.01) (0.91) = 0.0091.

Thus, P (D) = 0.0171 + 0.0091 = 0.0262.

Thus, the probability that a molding from the early morning run is defective, is 0.0262.

b.

The required probability is, P (F | D).

From the properties of the conditional distribution:

P (F | D)

= P (F ∩ D) / P (D)

= P (D ∩ F) / P (D)

= 0.0171/0.0262

= 0.6527.

Thus, the probability that the foreman forgot to shut off the machine the previous night, given that the plant manager randomly selects a molding from the early morning run and discovers it is defective, is 0.6527.