= 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 $m$, is
$$
\frac13q+(1-q)=\frac37,
$$
so $q=6/7$. 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 gives
$$
\frac67=q=\frac{(3/4)p}{(p+2)/4}=\frac{3p}{p+2}.
$$
Solving $6(p+2)=21p$ 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.
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