Health Tests

Solution

Answer: 22.9%

Let \(H\) denote the event that a person has HIV, and \(T^+\) denote the event that the test is positive. The prevalence of HIV in the population is \(0.3\%\), so \[ P(H) = 0.003, \qquad P(H^c) = 0.997. \] The test is \(99\%\) accurate in both directions. This means \[ P(T^+ \mid H) = 0.99 \] and \[ P(T^- \mid H^c) = 0.99. \] Therefore the false positive rate is \[ P(T^+ \mid H^c) = 0.01. \] We want the probability that someone actually has HIV given that they tested positive: \[ P(H \mid T^+). \] Using Bayes' theorem, \[ P(H \mid T^+) = \frac{P(T^+ \mid H)P(H)} {P(T^+ \mid H)P(H) + P(T^+ \mid H^c)P(H^c)}. \] \[ P(H \mid T^+) = \frac{(0.99)(0.003)} {(0.99)(0.003) + (0.01)(0.997)}. \] \[ P(H \mid T^+) \approx 0.229. \]