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?

Events

A = widget was produced by the old machine

B = widget was produced by the new machine.

D = widget was defective

Recall definition of conditional probability

P (X|Y) = P (X n Y) / P (Y) [ n represents set intersection operator ]

By the law of total probability,

P (A) = 1 / (1+3) = 1/4

P (B) = 3 / (1+3) = 3/4

Given:

Defective rate of new machine

P (D|B) = P (D n B) / P (B) = 0.08 = > P (D n B) = 0.08*0.75=0.0600

Since P (B) = P (D n B) + P (~D n B) = 0.75, this means that

P (~D n B) = 0.75-0.0600 = 0.69

Proceed to calculate probability of non-defective widget given it is produced by the new machine:

P (~D|B) = P (~D n B) / P (B) = 0.69 / 0.75 = 0.92

Conclusion

The probability that the widget is not defective given that it was produced by the new machine is 0.92.