Answer to Question #154618 in Statistics and Probability for asia

Question

Each item coming off a given production line is inpected by either inspector 1 or inspector 2. Inspector 1 inspects about 60% of the production items, while inspector 2 the rest. Inspector 1, who has been at her present job for some time, will not find 1% of the defective items she inspects. Inspector 2, who newer on the job, misses about 5% of the defective items he inspects. If an item that has passed an ispector is found to be defective, what is the probability that it was inspected by inspector 1?

Answer 1

P(1)=0.6; P(2)=0.4

P(D|1)=0.01; P(D|2)=0.05

Calculate P(1|D)

From Bayes theorem;

$P(1|D)=\frac{P(D|1).P(1)}{P(D|1).P(1)+P(D|2).P(2)}$

=$\frac{0.01*0.6}{0.01*0.6+0.05*0.4}$

=0.2308