 Solution

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.