In a certain town, 2% of the population suffers from disease D. A diagnostic test for D has sensitivity 90% (it reports positive for a diseased person) and specificity 85% (it reports negative for a healthy person).\n(i) If a randomly selected person tests positive once, what is the probability that he actually has the disease?\n(ii) If the same person undergoes two independent tests and both return positive, what is the updated probability that he has the disease?

Step-by-step solution

1. Let $D$ be the event that a person has the disease, $+$ the event of a positive test.
2. Given: $P(D)=0.02$, $P(+\mid D)=0.90$, $P(-\mid D^{\prime })=0.85\Rightarrow P(+\mid D^{\prime })=0.15$.
3. (i) Apply Bayes theorem: $P(D\mid +)=\frac{P(+\mid D)P(D)}{P(+\mid D)P(D)+P(+\mid D^{\prime })P(D^{\prime })}=\frac{0.90\times 0.02}{0.90\times 0.02+0.15\times 0.98}$.
4. Compute numerator $=0.018$, denominator $=0.018+0.147=0.165$, so $P(D\mid +)=0.018/0.165=0.1091\approx 0.109$.
5. (ii) For two independent tests both positive, use $P(++\mid D)=0.90^2$ and $P(++\mid D^{\prime })=0.15^2$.
6. Bayes again: $P(D\mid ++)=\frac{0.90^2\times 0.02}{0.90^2\times 0.02+0.15^2\times 0.98}$.
7. Numerator $=0.81\times 0.02=0.0162$; denominator $=0.0162+0.0225\times 0.98=0.0162+0.02205=0.03825$.
8. Thus $P(D\mid ++)=0.0162/0.03825=0.4235\approx 0.424$.

Concept tested: Bayes theorem with multiple tests

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