Question 1088140: A factory runs three shifts. The first two shifts produce 500 widgets per week each, while the third shift produces 350. Quality control had determined that 10% of the first shifts production is defective, while the second and third shifts have defective rates of 7% and 5%, respectively.
a. Interpret this information as conditional probabilities
b. What is the probability that a randomly selected widget from the factorys overall production will be defective?
c. A randomly selected widget is found to be defective. What is the probability that the widget came from the first shift?
I dont know if I use Bayes Theorm, P(ABC)= P(B/A)xP(A)+P(C/B)xP(B) or what??
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! For the first, 50 of 500 are defective. P(D|FS}=0.1
For the second, 35 of 500 are defective
For the third, 17.5 of 350 are defective
this is 102.5 out of 1350 or 0.0759.
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For the second part, I make a contingency table
========Def=====ND=======Total
First---50------450-------500
Second--35------465-------500
Third---17.5----332.5-----350
Total---102.5---1247.5----1350
If defective, it is 50/102.5 or 0.488 probability it came from first shift.
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