= Solution
For a stop-phase arrival at time $t$, the delay is $s=r-t$. An interval of delays of width $ds$ therefore corresponds to an interval of arrival phases of the same width, because $|dt/ds|=1$. Its unconditional <probability> is $ds/R$. Hence the positive-delay part has density
$$
\boxed{p(s)=\begin{cases}1/R,&0<s\leq r,\\0,&r<s\leq R.\end{cases}}
$$
This ordinary density integrates to $r/R$, not to one: the remaining <probability mass> is at $s=0$. Thus $S$ has a <uniform distribution with an atom at zero>, with mixing weight $q=r/R$. Conditional on a positive delay, the density would instead be $1/r$ on $(0,r)$.
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