Question

Suppose 0.5% of children under age 5 have lead poisoning. A blood test for lead poisoning is known to have a false positive rate of 5% and a false negative rate of 2%. If a child is tested and the result indicated lead poisoning, what is the probability that the child actually has lead poisoning?

Answer 1

Given,

P(Poisoning) = 0.005

So P(No Poisoning) = 1 - 0.005 = 0.995

P(Positive | No Poisoning) = 0.05

P(Negative | Poisoning) = 0.02

So P(Positive | Poisoning) = 1 - 0.02 = 0.98

Hence by Baye's theorem:

P(Poisoning | Positive)

$= \frac{0.005*0.98}{0.005*0.98 + 0.995*0.05}$

= 0.0897