Question

• Suppose that 1 out of every 200 students has Meningitis. Suppose there is a test T for Meningitis. Suppose that this test, like any test, is NOT 100% accurate. In particular, sup- pose that the probability that you test positive GIVEN that you have Meningitis equals .97. Suppose that the probability that you test negative GIVEN that you DO NOT have Mengingitis equals .92. Suppose you pick a random student, and test them for Meningitis using the test T.

  Consider the following associated events:

  – Event A: “Did the student actually have Meningitis?”
  – Event B: “Did the student actually NOT have Meningitis?”– Event C: “Did the student test positive?”
  – Event D: “Did the student test negative?”

  Compute P(A|C).

Answer 1

Given,

P(A) = 1/200 = 0.005

P(C | A) = 0.97

P(D | B) = 0.92

P(B) = 1 - 0.005 = 0.995

P(C | B) = 1 - P(D | B) = 1 - 0.92 = 0.08

By law of total probability,

P(C) = P(A) P(C | A) + P(B) P(C | B)

= 0.005 * 0.97 + 0.995 * 0.08 = 0.08445

By Bayes theorem,

P(A|C). = P(C | A) P(A) / P(C)

= 0.97 * 0.005 / 0.08445

= 0.05743