Question

A company has two machines that produce widgets. The new machine produces 75% widgets and the older machine produces 25% widgets. Further, the new machine produces 10% defective widgets, while the older machine produces 30% defective widgets. If a widget is randomly selected what is the probability that it is defective?

Answer 1

A company has two machines that produce widgets. The new machine produces 75% widgets and the old one produces 25% widgets.

The new machine produces 10% defective widgets, and the old machine produces 30% defective widgets.

So, given that

P(A randomly selected widget is produced by new machine)=75/100=0.75

P(A randomly selected widget is produced by old machine)=25/100=0.25

P(widget is defective|produced by new machine)=10/100=0.10

P(widget is defective|produced by old machine)=30/100=0.30

To find the chance that a randomly slected machine is defective.

So, we have to find

P(A randomly selcted machine is defective)

=P(A machine is defective and produced by old machine)+P(A machine is defective and produced by new machine)

From conditional probability, we know that P(A and B)=P(A)P(B|A).

Now, by total probability theorem, this becomes

=[P(widget is produced by new machine)*P(widget is defective|produced by new machine)]+[P(widget is produced by old machine)*P(widget is defective|produced by old machine)]

=0.75*0.10+0.25*0.30

=0.075+0.075

=0.150.

So, the probability that a randomly selected machine is defective is 0.150.