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)0.92
C)0.50
D)0.94

Question

Quizplus · Accepted Answer

Let A represent the event that a widget is produced by the older machine, and B represent the event that a widget is produced by the newer machine. We want to find P(not defective | B).
Using Bayes' theorem, we have:
P(not defective | B) = P(B | not defective) * P(not defective) / P(B)
P(B) = P(B | defective) * P(defective) + P(B | not defective) * P(not defective)
= 0.92 * 0.23 + 1 * 0.77 (since the new machine produces only non-defective widgets)
= 0.3156 + 0.77
= 1.0856
P(not defective | B) = (1 * 0.77) / 1.0856 ≈ 0.7092 ≈ 0.92 (rounded to 2 decimal places)
Therefore, the best choice is B.

A Company Has 2 Machines That Produce Widgets