# "A certain disease has an incidence rate of 2%. If

"A certain disease has an incidence rate of 2%. If the false negative rate is 10% and the false positive rate is 1%, compute the probability that a person who tests positive actually has the disease."
Why do we need to use Bayes' Theorem for this question?
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Superina0xb4i
A false positive is when the test is positive even though the person doesn't actually have the disease. The complement probability of a false positive is a true negative (i.e.: the test is negative and the person doesn't have the disease).
- 1% (the false positive rate) represents the probability that the test is incorrect given that the person doesn't have the disease.
- 99% (the true negative rate) represents the probability that the test is correct given that the person doesn't have the disease.
Neither of these two probabilities are what we want, since we don't know whether or not the person has the disease. All we know is that they tested positive (it could be a true positive or a false positive).
In general, there are four possibilities:
- True Positive: They had the disease (0.02) and the test was positive (0.9).
- False Negative: They had the disease (0.02) but the test was negative (0.1).
- False Positive: They didn't have the disease (0.98) but the test was positive (0.01).
- True Negative: They didn't have the disease (0.98) and the test was negative (0.99).
If a test is positive, then it could be a True Positive or a False Positive. We want to find the probability that, out of these two options, it was a True Positive. This yields:
$\frac{Pr\left[\mathsf{\text{True Positive}}\right]}{Pr\left[\mathsf{\text{True Positive or False Positive}}\right]}=\frac{\left(0.02\right)\left(0.9\right)}{\left(0.02\right)\left(0.9\right)+\left(0.98\right)\left(0.01\right)}=\frac{90}{139}=0.6474\dots$