probability question

Question

A certain disease has an incidence rate of 0.8%. If the false negative rate is 4% and the false positive rate is 2%, compute the probability that a person who tests positive actually has the disease.

Answer 1

We can use a tree diagram with 4 branches as follows:

D: people that have disease

D^C: people who do NOT have disease

P: people who test positive

P^C: people who test negative

D^C (.992) --------- P^C (.98) = .972

---------- P (.02) = .020 (false positives)

D (.008) ---------- P^C (.04) = .00032 (false negatives)

----------- P (.96) = .00768

Prob (D | P) = .00768 / .02768 = .277 ~ 27.7 % chance of having the disease given that they tested postive.

(This illustrates the phenomenon that, even though the tests seem fairly accurate, because most of the people testing are disease-free (the incidence rate is so low), most of the people who test positive are actually disease-free.)