Bayes' Law Practice Problem

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Two percent of men who participate in Covid 19 screening have the disease. Ninety-two percent of those with the disease test positive. Six percent of the men who do not have Covid 19 will also have positive test results. Given that a man has a positive test result, what is the probability he also has the disease?

My approach is this:

$P(T+|D) = 0.92$

$P(T+|D') = 0.06$

$P(D) = 0.02 -> P(D') = 0.98$

$P(D|T+) =$ $\Large\frac{P(T+|D) P(D)}{P(T+|D) P(D) + P(T+|D') P(D')}$

= $\Large\frac{0.92* 0.02}{0.92 * 0.02 + 0.06 * 0.98}$

Does this seem correct?