Probability exercise

Question

Suppose a blood test for a certain contagious disease is such that the probabilities of a false negative and a false positive are both equal to 0.01. Suppose further that the probability that a randomly chosen person is infected is equal to p. For what values of p, are more than 90% of the people for whom the test is positive infected?

Answer 1

Suppose you have a group of N people

Then Np are infected and N (1-p) not infected

Infected group Np x 0.99 test positive and Np x 0.01 test negative

Not infected N (1-p) x 0.01 test positive and N (1-p) x 0.99 test negative

Number testing positive : Np x 0.99 + N(1-p) x 0.01 = N (0.99p + 0.01 - 0.01 p) = N (0.98p + 0.01)

Of these Np x 0.99 are infected

So we want Np x 0.99 / N (0.98p + 0.01) > 0.90

0.99p > 0.882p + 0.009

0.108 p > 0.009

p > 0.083