Use base rates before interpreting a positive result

Let D denote disease and P a positive result. The initial branches are $P(D)=1/200$ and $P(D')=199/200$. Given D, positive/negative probabilities are $9/10$ and $1/10$. Given no disease, they are $2/100$ and $98/100$. The two outgoing probabilities at each branching point add to 1. The source’s teaching note identifies tree diagrams and conditional probability, referring to Lecture Example 8.

$P(D\cap P)=(1/200)(9/10)=9/2000=0.0045$.

$P(P')=(1/200)(1/10)+(199/200)(98/100)=2439/2500=0.9756$.

$P(D\mid P)=P(D\cap P)/P(P)=0.0045/[1-P(P')]=0.0045/(1-0.9756)=45/244$. The denominator includes positives from both disease-status groups; the test’s 0.9 rate is not this posterior probability.

Let P₁ and P₂ mean positive results on the first and second tests. Only P₁-positive people are retested. The source reuses 0.9 and 0.02 on the second test, assuming the test results are independent conditional on disease status and that these status-specific rates stay the same. This is an extra modelling assumption; one-test accuracy alone does not establish it.

Under that assumption,

$P(D\mid P_1\cap P_2)=\frac{(1/200)(9/10)(9/10)}{(1/200)(9/10)(9/10)+(199/200)(2/100)(2/100)}=\frac{2025}{2224}\approx0.911\quad(3\text{ s.f.}).$

The source writes $P(D\mid P_2)$ in the recalled group; its numerator/denominator and tree show that both tests are positive. The full conditioning event is made explicit here.