Let D be the event that an item is defective, and N be the event that an item is not defective. So, P(D)=0.3 and P(N)=1−0.3=0.7 Probability of an accurate result =
8
10
=0.8=P(RD∕D) Probability of an inaccurate result =1−0.8=0.2=P(RN∕D) So, P(D∩RN)=P(RN∕D)×P(D) =0.2×0.3=0.06 P(N∩RN)=P(RN∕N)×P(N) =0.8×0.7=0.56 0,P(RN)=P(D∩RN)+P(N∩RN) =0.06+0.56=0.62 P(D∕RN)=
P(D∩RN)
P(RN)
=
0.06
0.62
=
6
62
=
3
31
∴ The probability of an item is actually defective given device reports it as not defective is
