Question

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 randomly chosen widget was tested and was found to be defective, what is the probability that it was produced by the new machine?

Answer 1

The required probability of P(N/D) is 0.511.

Step-by-step explanation:

We are given that 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.

Let the probability that older machine produces widgets = P(O) = $\frac{1}{4}$ = 0.25

So, the probability that new machine produces widgets = P(N) = $\frac{3}{4}$ = 0.75

Also, let D = event that widgets produced are defective

So, the probability that defective widgets are produced by older machine = P(D/O) = 0.23

The probability that defective widgets are produced by new machine = P(D/N)= 0.08

Now, given that a randomly chosen widget was tested and was found to be defective, the probability that it was produced by the new machine is given by = P(N/D)

We will use the concept of Bayes' Theorem here to calculate the above probability, i.e;

       P(N/D)  =  $\frac{P(N) \times P(D/N)}{P(O) \times P(D/O)+P(N) \times P(D/N)}$

                    =  $\frac{0.75 \times 0.08}{0.25 \times 0.23+0.75 \times 0.08}$

                    =  $\frac{0.06}{0.1175}$

                    =  0.511