This stumped me, but I should have known better. I ended up guessing because my P (A | B) wasn’t working out. FYI A company has two machines that produce widgets. An older machine produces 16% defective widgets, while the new machine produces only 8% defective widgets. In addition, the new machine employs a superior production process such that it produces three times as many widgets as the older machine does. Given that a widget was produced by the new machine, what is the probability it is NOT defective? A) 0.06. B) 0.50. C) 0.76. D) 0.92. Your answer: B was incorrect. The correct answer was D) 0.92. The problem is just asking for the conditional probability of a defective widget given that it was produced by the new machine. Since the widget was produced by the new machine and not selected from the output randomly (if randomly selected, you would not know which machine produced the widget), we know there is an 8% chance it is defective. Hence, the probability it is not defective is the complement, 1 – 8% = 92%.
Just don’t let the word “given” fool you. Most of the time given is used when refering to a joint probability question but be careful
What’s tricky about this question? Seems very straight-forward to me.
I assumed this exactly: Since the widget was produced by the new machine and not selected from the output randomly (if randomly selected, you would not know which machine produced the widget), we know there is an 8% chance it is defective
B - new machine, notB - old machine, A - defective, notA-not defective you are given P(A|B), P(A|notB) and then you are asked P(notA|B)= 1-P(A|B). What wasn’t working out with conditional probabilities?
The info on the old machine is just a distractor. So, considering only the widgets produced by the new machine, how about the common sense, rather than formulas? If you know that 8% of the widgets produced by the newmachine are defect, what is the probability that yours is not defect? 92%
The two event are independent , given is of not help. Just consider the production of the new machine as one independent event. And the production of the old as another. In that case you can user the total propbility for the new machine: P(defect)+p(Nondefect)= 1 which implies that p(Nondefect)= 1-p(defect)= 1-0.08=0.92