5. (1 pt) Library/UVA-Stat/setStat212-Homework04/stat212-HW04- 11.p9 A foreman f

Question

5. (1 pt) Library/UVA-Stat/setStat212-Homework04/stat212-HW04- 11.p9 A foreman for an injection-molding firm admits that on 41% of his shifts, he forgets to shut off the injection machine on his line. This causes the machine to overheat, increasing the proba- bility that a defective molding will be produced during the early morning run from 4% to 22%. The plant manager randomly se- lects 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?

Explanation / Answer

Solution:

Given:
A foreman forgets to shut off machine 41% of the time.
If he forgets, 22% of moulds are defective.
If he does not, 4% of moulds are defective.
Given that a mould is defective, find probability that he forgot to turn off machine the night before.

Solution:

Define events
1. F = event of forgetting to shut off machine.
2. D = event of product being defective.

P(F and D) = 0.41*0.22 = 0.0902
P(F and ~D) = 0.41*(1-0.22) = 0.3198
P(~F and D) = (1-0.41)*0.04 = 0.0236
P(~F and ~D) = (1-0.41)*(1-0.04) = 0.5664

By definition of conditional probability
P(F|D)=P(F and D)/P(D)
=P(F and D)/[P(F and D)+P(~F and D)]
= 0.0902/(0.0902+0.0236)
= 0.7926 (approx.)

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