Question

A certain disease has an incidence rate of 0.2%. If the false negative rate is 5% and the false positive rate is 2%, compute the probability that a person who tests positive actually has the disease.

Answer 1

P(I)=Incidence rate of a disease= 0.2 %

P(F .N) =False negative rate = 5%

P (F.P)=False positive rate= 2%

Probability that a person who tests positive actually has the disease

                         =  $\frac{P(F.P)\times P(I)}{P(F.P)\times P(I)+P(F.N)\times P(I)}$

                 =  $\frac{.2\times2}{.2\times2+.2\times5}$

                 =  $\frac{.4}{1.4}$

                 = $\frac{2}{7}$