Shea Stuart

2022-07-01

Consider a disease that is on average caught by $1$ in $N$ people. Let's say that a disease test gives a false positive for $p\mathrm{%}$ of healthy people. For diseased people, it always gives a correct, positive answer.
What's the chance that a person who got a positive result is healthy?

Leslie Rollins

Imagine $1000N$ people. $\frac{1000N}{N}=1000$ of those people have the disease, $1000N-1000=1000\left(N-1\right)$ do not. Of the $1000\left(N-1\right)$ people who do not have the disease the test falsely reports that $1000p\left(N-1\right)$ do have the disease. The test also reports all $1000$ people who have the disease so reports a total of $1000+1000p\left(N-1\right)=1000\left(1+p\left(N-1\right)\right)$ people as having the disease. Of the $1000\left(1+p\left(N-1\right)\right)$ people reported to have the disease, $1000p\left(N-1\right)$ are actually healthy. The probability that a person reported to have the disease is actually healthy is $\frac{1000p\left(N-1\right)}{1000\left(1+p\left(N-1\right)\right)}=\frac{p\left(N-1\right)}{\left(1+p\left(N-1\right)\right)}$.