Question

A certain disease has an incidence rate of 0.4%. If the false negative rate is 6% and the false positive rate is 4%, compute the probability that a person who tests positive actually has the disease.

Give your answer accurate to at least 3 decimal places

Answer 1

Let's start by defining the events involved:
D= the event that a person has the disease
D'= the event that a person does not have the disease
T+ - the event that a person tests positive
T-= the event that a person tests negative
We are given that the incidence rate of the disease is 0.4%, which means P(D) = 0.004. We
are also given the false negative rate, which means P(T-ID) = 0.06, and the false positive rate,
which means P(T+|D') = 0.04
We want to compute the probability that a person who tests positive actually has the disease, which is given by Bayes' theorem:
P(DIT+) = P(T+|D)P(D) / [P(T+|D)P(D) + P(T+[D')P(D)]
To compute this probability, we need to compute P(T+|D) and P(T+|D'):

P(T+ID) = 1 - P(T-|D) = 1 - 0.06 = 0.94
P(T+|D') = false positive rate = 0.04
Now we can substitute these values into Bayes' theorem:
P(DIT+) = (0.94)(0.004) / [(0.94)(0.004) + (0.04)(1 - 0.004)]
=0.0853
Therefore, the probability that a person who tests positive actually has the disease is 0.0853 or approximately 0.085 to 3 decimal places