Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 2 / 11

Every set of $S$ students is equally likely to be the specified college. There are $\left(\genfrac{}{}{0ex}{}{N}{S}\right)$ choices in total and $\left(\genfrac{}{}{0ex}{}{L}{S}\right)$ choices consisting entirely of the specified group. Therefore the hypergeometric distribution gives

Choose $n$ students from the $L$ lazy students and $S-n$ from the other $N-L$. Dividing by the total number of possible college memberships givesThe assumptions ensure all these values are possible. This sampling is without replacement, as required by the hypergeometric distribution.

By symmetry, each student in the university has probability $(S/N)(1/S)=1/N$ of being in the college and then selected. Hence the prior probability is

The lecture attendances are independent Bernoulli trials with success probability $f$. Their count has the binomial distribution:

Here “absent” denotes absence at both observed lectures. The assumed independence makes its conditional probability

For attendance probability $9/10$, absence on both independent observations has conditional probability

The law of total probability over the two student groups gives

The prior probability is $1/10$. Using Bayes' theorem, the posterior probability after two absences isThus the probability rises from 10% to about 73.5% within the assumed attendance model. Two absences are 25 times as likely under the lazy model, overcoming the prior odds of $1:9$; they still do not make the classification certain.