Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 4

The exponential series proves normalization of a probability distribution:Let $M$ and $F$ count the two labels. Conditional on $N=M+F=k+l$, independent labeling gives a binomial distribution, so the joint probability mass function isEach bracket is a normalized Poisson distribution. Summing over $l\ge 0$ therefore givesThe factorization also proves that $M$ and $F$ are independent random variables. This is Poisson thinning; the phrase about ignoring the other count refers to taking a marginal distribution, rather than fixing that count.

Let $D$, $C$, and $V_0$ denote the three disjoint categories, and $+$ or $-$ the test result. The law of total probability givesTo calculate the requested conditional probability, first combine the two categories with the virus. Their joint negative-test probability is $0.1(0.05)+0.2(0.1)=0.025$, whereas $\mathbb{P}(-)=1-0.31=0.69$. Bayes' theorem therefore givesThese numbers use the specified population proportions and test probabilities; the negative result reduces the initial combined proportion $0.3$ to about $3.62\%$ within this model.