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 widget was produced by the new machine, what is the probability it is not defective?

Answer 1

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.