Question

A test for a certain disease provides a correct diagnosis (gives a positive result, when the patient has the disease) with probability 0.97, and wrongly diagnoses (gives a positive result, when the patient doesn’t have the disease) with probability 0.02. It is known that the disease is rare: published figures indicate it occurs in only 3 per 1000 individuals. Suppose that a randomly chosen individual is given the test. Calculate the probability (to at least four decimal places) that:

(a) The test is positive.

(b) The individual actually suffers from the disease, given that the test is positive.

Answer 1

Let A = person has the disease

Let B = positive test

Given ,

P(A) = 3 / 1000 = 0.003

P( not A ) = 1 - 0.003 = 0.997

P(B | not A) = 0.02

P(B | A) = 0.97

a)   The probability that the test is positive :

P(B) = P(B | A). P(A) + P(B | not A). P( not A )

  = 0.97 *0.003 + 0.02 * 0.997

= 0.0029 + 0.0199

= 0.0228

b) The probability that the individual actually suffers from the disease, given that the test is positive.

We wish to calculate P(A| B) , using Bayes’s theorem, we can compute this probability.

P(BA)P(A) PAB) = P BLAPCA) + PB not AP(nota) PBAPA) P(B)

  

0.97 +0.003 00228

  

= 0.1276