Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 8 / c / iii

Let $M$ count the ringed birds among the two captures, and let $D$ be the event that the detector records at least one ring. The hypergeometric distribution gives the prior probabilitiesUnder the usual additional assumption of independent per-ring detection, with no false records, the likelihoods $\mathbb{P}(D\mid M=m)$ are $0$, $9/10$, and $1-(1/10{)}^2=99/100$. Therefore $\mathbb{P}(D)=12/25+33/250=153/250$. Applying Bayes' theorem yields $\mathbb{P}(M=0\mid D)=0,\mathbb{P}(M=1\mid D)=\frac{40}{51},\mathbb{P}(M=2\mid D)=\frac{11}{51}$. Storing only the last detected number means that a stored record does not identify how many rings were present. By label symmetry, the identity of that stored ring does not further alter this distribution.

The text specifies each ring's marginal detection probability, but does not explicitly specify independence of the detection errors. To keep this modelling qualification visible, let $\ell =\mathbb{P}(D\mid M=2)$. The marginal detection rate only implies $9/10\le \ell \le 1$. Without the extra independence assumption, the full family of possible posterior distributions isIn particular $1/5\le \mathbb{P}(M=2\mid D)\le 5/23$. The displayed numerical answer corresponds to $\ell =99/100$; the general answer exposes exactly what the detector's marginal specification alone establishes.