A company has two 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 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) 0.92
D) 0.94

Question

Quizplus · Accepted Answer

Let us use Bayes' theorem to find the probability that the widget produced by the new machine is not defective.

Let A be the event that a widget is produced by the new machine, and B be the event that the widget is not defective.

We want to find P(B|A), i.e., the probability that the widget is not defective, given that it was produced by the new machine.

Bayes' theorem tells us that:

P(B|A) = P(A|B) * P(B) / P(A)

We know that P(A|B) is the probability that the new machine produced the widget given that it is not defective. In other words, we know that P(A|B) = 1, since the problem states that the new machine produces only 8% defective widgets.

We also know that P(B) is the overall probability of a widget being not defective. This is given by:

P(B) = P(B|old) * P(old) + P(B|new) * P(new)

where P(B|old) is the probability of a widget being not defective, given that it was produced by the old machine (which is 1 - 0.23 = 0.77), P(old) is the probability of the widget being produced by the old machine (which is 1/4, since the new machine produces three times as many widgets), P(B|new) is the probability of a widget being not defective, given that it was produced by the new machine (which is 1 - 0.08 = 0.92), and P(new) is the probability of the widget being produced by the new machine (which is 3/4).

So we have:

P(B) = 0.77 * 1/4 + 0.92 * 3/4 = 0.8875

Finally, we need to find P(A), the probability that the widget was produced by the new machine. This is also given by:

P(A) = P(A|old) * P(old) + P(A|new) * P(new)

where P(A|old) is the probability of the widget being produced by the new machine, given that it was produced by the old machine (which is 0, since the old machine cannot produce widgets from the new machine), and P(A|new) is the probability of the widget being produced by the new machine, given that it was produced by the new machine (which is 1).

So we have:

P(A) = 0 * 1/4 + 1 * 3/4 = 0.75

Now we can plug in everything we know into Bayes' theorem:

P(B|A) = P(A|B) * P(B) / P(A)

= 1 * 0.8875 / 0.75

= 0.94

Therefore, the probability that a widget produced by the new machine is not defective is 0.94.

A Company Has Two Machines That Produce Widgets