A certain disease has an incidence rate of 0.1%. If the false negative rate is 8% and the false positive rate is 3%, compute the probability that a person who tests
positive actually has the disease.

0.31 or 31%

Let A be the event that the disease is present in a particular person

Let B be the event that a person tests positive for the disease

The problem asks to find P(A|B), where

P(A|B) = P(B|A)*P(A) / P(B) = (P(B|A)*P(A)) / (P(B|A)*P(A) + P(B|~A)*P(~A))

In other words, the problem asks for the probability that a positive test result will be a true positive.  

P(B|A) = 1-0.02 = 0.98 (person tests positive given that they have the disease)

P(A) = 0.009 (probability the disease is present in any particular person)

P(B|~A) = 0.02 (probability a person tests positive given they do not have the disease)

P(~A) = 1-0.009 = 0.991 (probability a particular person does not have the disease)

P(A|B) = (0.98*0.009) / (0.98*0.009 + 0.02*0.991)

= 0.00882 / 0.02864 = 0.30796

*round however you need to but i am leaving it at 0.31 or 31%*

If you found this helpful please mark brainliest

answer: -8.2562234e+13