Using Bayes’ Theorem to Find the Probability of Disease Given a Positive Test

In a certain city, 1\% of the population has a rare disease. A diagnostic test for the disease gives a true positive rate of 95\% (the probability that the test is positive if the person has the disease) and a false positive rate of 5\% (the probability that the test is positive if the person does not have the disease). If a person tests positive, what is the probability that they actually have the disease? (Hint: Use Bayes’ Theorem to solve.) $P(D) = 0.01$ $P(\text{Positive} | D) = 0.95$ $P(\text{Positive} | \neg D) = 0.05$ $P(\neg D) = 0.99$ $P(D | \text{Positive}) = \frac{P(\text{Positive} | D) \cdot P(D)}{P(\text{Positive})}$ Where: $P(\text{Positive}) = P(\text{Positive} | D) \cdot P(D) + P(\text{Positive} | \neg D) \cdot P(\neg D)$ $= 0.95 \times 0.01 + 0.05 \times 0.99 = 0.0095 + 0.0495 = 0.059$ $P(D | \text{Positive}) = \frac{0.95 \times 0.01}{0.059} \approx 0.161$