Answer to Question #218451 in Statistics and Probability for Ernest

Question

A blood test indicates the presence of a particular disease 95% of the time when the disease is actually present. The same test indicates the presence of the disease 0.5% of the time when the disease is not present. One 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

Y = positive test result

D = disease is present (and ~D = not D)

Using Baye’s theorem:

$P(D|Y) = \frac{P(Y|D)P(D)}{P(Y|D)P(D)+P(Y|~D)P(~D)} \\ = \frac{0.95 \times 0.01}{(0.95 \times 0.01)+(0.005\times 0.99)} \\ = 0.657$