Question

A blood test indicates the presence of a particular disease 96​% of the time when the disease is actually present. The same test indicates the presence of the disease 0.8​% of the time when the disease is not present. Two percent of the population actually has the disease. Calculate the probability that a person has the​ disease, given that the test indicates the presence of the disease.

Answer 1

P(actually has the disease) = 0.02

P(test indicates the presence of the disease | actually has the disease) = 0.96

P(test indicates the presence of the disease | the disease is not present) = 0.008

P(test indicates the presence of the disease) = P(test indicates the presence of the disease | actually has the disease) * P(actually has the disease) + P(test indicates the presence of the disease | the disease is not present) * P(the disease is not present)

                = 0.96 * 0.02 + 0.008 * (1 - 0.02)

                = 0.02704

P(person has the​ disease | the test indicates the presence of the disease) = P(test indicates the presence of the disease | actually has the disease) * P(actually has the disease) / P(test indicates the presence of the disease)

              = 0.96 * 0.02 / 0.02704

              = 0.71