Question

You are a research engineer working for a pharmaceutical company that is conducting clinical trials for a new disease screening test.  Suppose that 2% of a clinic’s patients are known to have this disease. The test you developed is positive in 98% of the patients with the disease, but also positive in 3% of patients who do not have the disease. If a person who is chosen at random from the clinics patients is given the test and it comes out positive, what is the probability that the person actually has the disease.

HINT: Consider the fact that of the patient actually having the disease, and whether the test shows he/she has the disease, as independent events.

Answer 1

We can form the following contingency table using the given information and assuming that the clinic has 100 patients

	Test positive	Test negative	Total
Has disease	0.98 * 2 = 1.96	2 - 1.96 = 0.04	2
No disease	0.03 * 98 = 2.94	98 - 2.94 = 95.06	100 - 2 = 98
Total	4.9	95.1	100

P(Has disease | Test positive) = P(Has disease and Test positive) / P(Test positive)

= (1.96/100) / (4.9/100)

= 0.40