A company has 2 machines that produce widgets.An older machine produces 23% defective widgets,while the new machine produces only 8% defective widgets.In addition,the new machine produces 3 times as many widgets as the older machine does.Given a randomly chosen widget was tested and found to be defective,what is the probability it was produced by the new machine?
A)0.511
B)0.08
C)0.15
D)0.489

Question

Quizplus · Accepted Answer

Let A be the event that the widget was produced by the older machine and B be the event that the widget was produced by the new machine. We need to find the probability of event B given that the widget is defective, i.e., P(B|D).
Using Bayes' theorem, we have:
P(B|D) = P(D|B) * P(B) / P(D)
where
P(D|B) = probability of a defective widget given that it was produced by the new machine = 0.08
P(B) = probability of a widget produced by the new machine = 3/4 (since the new machine produces 3 times as many widgets as the older machine)
P(D) = probability of a defective widget = P(D|A) * P(A) + P(D|B) * P(B)
where
P(D|A) = probability of a defective widget given that it was produced by the older machine = 0.23
P(A) = probability of a widget produced by the older machine = 1/4
Putting in the values, we get:
P(D) = 0.23 * 1/4 + 0.08 * 3/4 = 0.0925
Therefore, 
P(B|D) = 0.08 * 3/4 / 0.0925 = 0.511
Hence, the answer is A) 0.511.

A Company Has 2 Machines That Produce Widgets