Let there be two events $A$ and$$ $B$. Denote the probabilities of $A$ and $B$ occurring independently be $P(A)$ and $P(B)$. Let $P(A|B), P(B|A)$ denote the probability of event $A$ occurring given $B$ happens, and the probability of event $B$ occurring given $A$ happens. Bayes’ Theorem states,

$P(A|B)P(B)=P(B|A)P(A)=P(A\cap B)$.

Take an example where the chance of getting COVID-19 is 64%. For our COVID-19 test, there will be a 1% chance that the test obtains a false positive, and a 5% chance that the test obtains a false negative.

Question: Suppose the test returns positive. What is the probability you really have COVID-19?

$P(\text{COVID}|+)\\=\frac{P(+|\text{COVID})}{P(+)}P(\text{COVID})\\ =0.99\times0.64\div \\(P(+|\text{COVID})P(\text{COVID})+P(+|!\text{COVID})P(!\text{COVID}))\\=0.972...$