Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 17

Temporarily label the individual pieces. A random placement is a uniform injection of thirty-two distinct pieces into sixty-four squares, giving ${}_{64}{P}_{32}$ equally likely outcomes. For a colour-matching placement, the black pieces have ${}_{32}{P}_{16}$ available injections, and independently the white pieces have the same number within their disjoint set of squares. ThereforeHere ${}_n{P}_r=n!/(n-r)!$. Treating same-colour pieces as indistinguishable gives the same probability, because the common within-colour labeling factors cancel. The two colour choices in the favourable count occupy disjoint square sets; no independence assumption about unrestricted placements is required.

The lazy choice weights are $1/w$ and $1/(3w)$, in ratio three to one. Normalizing gives $\boxed{l=3/4}$. Let $Q$ denote selection of the first question and let $q=P(\text{Lazy}\mid Q)$. Its conditional expected value, divided by $m$, isso $q=6/7$. If the overall lazy fraction is $p$, the law of total probability gives $P(Q)=(3/4)p+(1/2)(1-p)=(p+2)/4$. Bayes' theorem then givesSolving $6(p+2)=21p$ yields $\boxed{p=4/5}$. The observed mean is conditioned on the chosen question, so it cannot be equated directly to the unconditional mean over all students.