Question

In a certain lot of personal computers, it is known that 1 % have some minor defect as they come off the production line. They are put through a test procedure, which detects any defect 98 % of the time if a defect is really present, and indicates a defect 1 % of the time even though there is none present. What is the probability that a. a computer will be classified defective as a result of the test procedure? b. a computer is in fact defective if the test indicates that it is defective?

Answer 1

let T=event of the computer detecting a defect

D=event of having a defect

D'=event of not having a defect

it is given in the problem

probability of having a defect=0.01

probability of not having a defect=1-P(D)=1-0.01=0.99

probability of the test detecting a defect when there is a defect=P(T|D)=0.98

probability of the test detecting a defect when there is no defect= P(T|D')=0.01

a)probability a computer will be classified defective as a result of the test procedure

=P(D)*P(T|D)+P(D')*P(T|D')=0.01*0.98+0.99*0.01 = 0.0197

b)probability of the computer is in fact defective if the test detects as defective

=P(D|T)

= , (bayes theorem)

=0.497