Separate infection counts from a later conditional-test model

Assume the infection probability is the same p for every sampled person and that infection indicators for the sampled people are independent.

Let $X\sim B(25,p)$. The equation is $P(X\leq2)=\sum_{k=0}^2\binom{25}kp^k(1-p)^{25-k}=0.0982$, equivalently $(1-p)^{25}+25p(1-p)^{24}+300p^2(1-p)^{23}=0.0982$. Solving gives $p\approx0.2000123866$, which is the printed $p=0.200$ (3 s.f.).

Let $D$ count samples with at most two infected people. Under independent sample outcomes, $D\sim B(30,0.0982)$. Then $P(D\geq5)=1-P(D\leq4)\approx0.1663901171=0.166$ (3 s.f.). Use the given .0982, not a probability recalculated from rounded p=.200.

The question now sets a new p=.10. Write I for infected and + for positive. The tree gives $P(+)=.10(.85)+.90(.04)=.121$. Bayes’ rule gives $P(I\mid+)=\dfrac{.10(.85)}{.121}=\dfrac{85}{121}\approx.7024793388=.702$ (3 s.f.).

$P(I^c\mid+)=1-P(I\mid+)=36/121\approx.2975206612=.298$ (3 s.f.). This is the fraction of positive results belonging to uninfected people in the given hypothetical model.

The source accepts a reasoned yes argument: about 70.2% of positive results identify infected people, and earlier identification could allow timely treatment or help prevent spread in the hypothetical setting.

The source also accepts a reasoned no argument: close to 30% of positive results belong to uninfected people, so responses to false positives could cause unnecessary treatment and resource use; grouping such people with infected people in quarantine could expose them to infection. These are conditional discussion arguments. The probabilities alone do not settle worthwhileness without the consequences, costs and use of the test results.