maths probability question

May 2016
391
3
england
hi i cant work this out

Its is estimated that 0.3% of a large population have a particular disease. a test developed to detected the disease gives a false positive in 4% of tests and a false negative in 1% of tests. a person is tested positive for the disease. what is the probability that the person actually has the disease?
 

romsek

MHF Helper
Nov 2013
6,738
3,034
California
$P[diseased|positive]P[positive] = P[positive|diseased]P[diseased]$

$P[diseased|positive] = \dfrac{P[positive|diseased]P[diseased]}{P[positive]}$

$P[positive|diseased] = 1-0.01 = 0.99$, this is 1 minus a false negative

$P[diseased]=0.3$

$P[positive|not diseased] = 0.04$, this is a false positive

$P[not diseased] = 1 - P[diseased] = 1 - 0.3 = 0.7$

$P[positive] = P[positive|diseased]P[diseased] + P[positive|not diseased]P[not diseased] = (1-0.01)(0.3) + (0.04)(1-0.3) = (0.99)(0.3)+(0.04)(0.7) = 0.325$


So

$P[diseased|positive] = \dfrac{(0.99)(0.3)}{0.325} = 0.913846 \approx 0.914$