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 that a widget was produced by the new machine,what is the probability it is not defective?
A)0.06
B)0.50
C)0.92
D)0.94

Question

Quizplus · Accepted Answer

Let A be the event that a widget is produced by the old machine, and B be the event that a widget is produced by the new machine. 
P(A) = 1/4 (since the new machine produces 3 times as many widgets as the old machine, the ratio of the number of widgets produced by the old machine to the total number of widgets produced is 1/4)
P(B) = 3/4
P(defective given A) = 0.23
P(defective given B) = 0.08
We want to find P(not defective given B), which is equivalent to 1 - P(defective given B).
Using Bayes' Theorem: 
P(defective given B) = P(B and defective) / P(B) = (0.08 x 3/4) / 1 = 0.06
Therefore, P(not defective given B) = 1 - P(defective given B) = 1 - 0.06 = 0.94. 
Hence the answer is (C).

A Company Has 2 Machines That Produce Widgets