Probability

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 ‌

3

31

.