In a town, $1\%$ of the population suffers from disease $D$. A diagnostic test for $D$ has the following properties: it correctly identifies a diseased person (positive result) $95\%$ of the time, and correctly identifies a healthy person (negative result) $90\%$ of the time. (i) If a randomly selected person tests positive once, what is the probability that the person actually has the disease? (ii) If the same person undergoes two independent tests and both return positive, what is the updated probability of having the disease?

Step-by-step solution

1. Let $D$ be the event "person has disease" and $+$ the event "test is positive".
2. Prior probability: $P(D)=0.01$, $P(D^c)=0.99$.
3. Test characteristics: $P(+|D)=0.95$ (sensitivity), $P(+|D^c)=1-0.90=0.10$ (false‑positive rate).
4. For one positive test, apply Bayes: $P(D|+)=\frac{P(+|D)P(D)}{P(+|D)P(D)+P(+|D^c)P(D^c)}$.
5. Compute numerator: $0.95\times 0.01=0.0095$.
6. Compute denominator: $0.0095+0.10\times 0.99=0.0095+0.099=0.1085$.
7. Thus $P(D|+)=\frac{0.0095}{0.1085}=0.0876$ (about $8.8\%$).
8. For two independent tests, the likelihoods multiply: $P(++|D)=0.95^2$, $P(++|D^c)=0.10^2$.
9. Apply Bayes again: $P(D|++)=\frac{0.95^2\times 0.01}{0.95^2\times 0.01+0.10^2\times 0.99}$.
10. Numerator: $0.9025\times 0.01=0.009025$.
11. Denominator: $0.009025+0.01\times 0.99=0.009025+0.0099=0.018925$.
12. Therefore $P(D|++)=\frac{0.009025}{0.018925}\approx 0.477$ (about $47.7\%$).

Concept tested: Bayes theorem with prior probabilities

Published 5 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked