In a town, 0.3% of the population has disease D. A diagnostic test has sensitivity 80% and specificity 95%. If a person undergoes two independent tests and both return positive, what is the probability that the person actually has disease D?

Step-by-step solution

1. Let $D$ be the event that the person has the disease. Given $P(D)=0.003$ and $P(\bar{D})=0.997$.
2. Sensitivity gives $P(\text{+}|D)=0.8$; specificity gives $P(\text{+}|\bar{D})=0.05$.
3. For two independent tests, $P(\text{++}|D)=0.8^2=0.64$ and $P(\text{++}|\bar{D})=0.05^2=0.0025$.
4. Apply Bayes’ theorem: $P(D|\text{++})=\frac{P(\text{++}|D)P(D)}{P(\text{++})}$, where $P(\text{++})=0.64\times 0.003+0.0025\times 0.997=0.0044125$.
5. Compute the numerator: $0.64\times 0.003=0.00192$.
6. Thus $P(D|\text{++})=\frac{0.00192}{0.0044125}\approx 0.4353$.
7. Express as an exact fraction: $\frac{0.00192}{0.0044125}=\frac{192}{4412.5}=\frac{384}{8825}$.

Concept tested: Bayes theorem with multiple tests

Published 18 September 2026 · How this solution was prepared and checked