# Test for HIV. There's a false positive rate of 0.025 and a false negative rate of 0.08. Let's say a particular patient has a probability of testing positive for HIV of 0.005. The patient gets tested and it's positive. What are the chances that the patient actually has HIV?

Test for HIV. There's a false positive rate of 0.025 and a false negative rate of 0.08. Let's say a particular patient has a probability of testing positive for HIV of 0.005. The patient gets tested and it's positive. What are the chances that the patient actually has HIV?
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Brendon Bentley
This is a typical application of Bayes formula,
$P\left(B|A\right)=P\left(A|B\right)P\left(B\right)/\left(P\left(A|B\right)P\left(B\right)+P\left(A|\overline{B}\right)P\left(\overline{B}\right)\right).$
Here $A$ is testing positive, $B$ is having HIV, and $\overline{B}$ is not having HIV. $P\left(A|B\right)$ is the true positive rate, the complement of the fpr $.08$. $P\left(A|\overline{B}\right)$ is the false positive rate $.025$. Substituting the numbers you give, $P\left(B|A\right)=.92\ast .005/\left(.92\ast .005+.025\ast .995\right)=0.1560645.$