Use Bayes' rule to find the indicated probability. The incidence of a certain disease on the island of Tukow is 4%. A new test has been developed to diagnose the disease. Using this test, 91% of those who have the disease test positive while 4% of those who do not have the disease test positive (false positive). If a person tests positive, what is the probability that he or she actually has the disease?

Accepted Solution

These are the events in the question above:

D - has disease 

H - healthy (does not have disease) 

P - tests positive 

It is the probability that a person has the disease AND tests positive divided by the probability that the person tests positive.

Sick, + [.04*.91] = .0364 

Sick, - [.04*.09] = .0036 

Healthy, + [.96*.04] = 0.0384

Healthy, - [.96*.96] = .9216 

.0364 / (.0364 + .0.0384) = 0.487