Conditional Probability (Diseases)

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A town with a population 10,000 has suffered an outbreak of dragon pox, with 3% of the population being infected. There is a test to diagnose the disease. If you have the dragon pox, the test will correctly register positive 97% of the time and will yield a false negative 3% of the time. If you do not have the dragon pox the test will register that correctly 97% of the time and will yield a false positive 3% of the time. Given that you tested positive for the disease, what is the probability that you actually have dragon pox?

I just do not get this question at all.

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Hint

Need Bayes Theorem.

Let $A =$ event that you have the disease.

Let $B =$ event that you tested positive.

Let $(AB) =$ event that both of the events $A$ and $B$ occurred.

Per wikipedia article

$$p(A|B) = \frac{p(AB)}{p(B)}.$$

$p(AB)$ can be calculated as chance of having disease $\times$ chance of getting a true positive.

$p(B)$ can be calculated as the above chance

plus the chance of not having disease $\times$ chance of getting false positive.