Distinguish finite sampling from independent-person conditioning

Choose three chocolate and five blueberry cupcakes from the finite inventory, without replacement: $P=\binom{15}3\binom{25}5/\binom{40}8\approx0.314$ (3 s.f.).

For the order CCCBBBBB, multiply the changing counts, then multiply by the $8!/(3!5!)$ colour orders:

$\frac{15}{40}\frac{14}{39}\frac{13}{38}\frac{25}{37}\frac{24}{36}\frac{23}{35}\frac{22}{34}\frac{21}{33}\frac{8!}{3!5!}\approx0.314\quad(3\text{ s.f.}).$

Vaccinated/unvaccinated probabilities are 0.85/0.15. The flu/no-flu branches are 0.23/0.77 and 0.91/0.09; the no-flu complements fill the unlabelled source branches. Hence $P(F)=(0.85)(0.23)+(0.15)(0.91)=0.332=83/250$.

Using the source’s independent-person model, exactly one vaccinated flu case can occur in two orders. Therefore

$P(\text{exactly one vaccinated}\mid\text{both flu})=\frac{(0.85)(0.23)(0.91)(0.15)\,2!}{(0.332)^2}\approx0.484\quad(3\text{ s.f.}).$

The denominator restricts both people to the flu population; it is not the unconditional vaccination probability.