suppose that only 2 of individuals undergoing a particular m

suppose that only 2% of individuals undergoing a particular medical test have the disease that the test is intended to identify. if an individual has the disease, the probability that the test indicates the presence of the disease is 0.98. if an individual does not have the disease, the probability that the test falsely gives a positive result for the disease is 0.5. if the test result for an individual is positive, what is the probability that the individual actually has the disease?

Solution

A:The event that the patient has the disease.

B:The event that the test gives true result.

P(A)=0.02

P(B|A)=0.98

P(B|A\')=0.5

P(B|A)=P(AB)/P(A)

P(AB)=P(B|A)*P(A)=0.98*0.02

P(A\'B)=P(B|A\')*P(A\')

=>P(B)-P(AB)=P(B|A\')*P(A\')

=>P(B)=P(AB)+P(A\')*P(B|A\')=0.98*0.02+0.5*0.98

Thus,P(A|B)=P(AB)/P(B)=0.3846

Thus,given the test result is positive,probability that the individual actually has the disease=0.3846