Question

People are being tested for a disease that 0.4% of the population has. The test is 98% successful; if the disease is present, it will be positive with a 0.98 probability and if the disease is not present, the test will be negative with a 0.98 probability.  If the test is positive what is the chance the individual has the disease?

Answer 1

Bayes' Theorem: P(A | B) = P(A & B)/P(B)

P(has disease) = 0.004

P(no disease) = 1 - 0.004 = 0.996

P(disease and positive) = 0.004x0.98 = 0.00392

P(no disease and positive) = 0.996x(1-0.98) = 0.01992

P(has disease | positive) = P(has disease and tested positive) / P(positive)

= 0.00392/(0.00392 + 0.01992)

= 0.1644