Question 997079
 
Given:
A foreman forgets to shut off machine 39% of the time.
If he forgets, 15% 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) = 0.39*0.15=0.0585
P(F and ~D) = 0.39*(1-0.15)=0.3315
P(~F and D) = (1-0.39)*0.04=0.0244
P(~F and ~D) = (1-0.39)*(1-0.04)=0.5856
 
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)]
=0.0585/(0.0585+0.0244)
=0.706 (approx.)
 
Answer:
The probability that the foreman forgot to shut the machine is 0.706, given that the product was defective.