Answer to Question #218226 in Statistics and Probability for Bless

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

Let Y is a positive test result, D - disease is present.

$P(D|Y)=\frac{p(Y|D)P(D)}{P(Y|D)P(D)+P(Y|notD)P(notD)}=$

$=\frac{0.95*0.01}{0.95*0.01+0.005*0.99}=\frac{0.0095}{0.01445}=0.6574.$