= Solution
Under the stated topology <probabilities>, $P(A)=2/3$ and $P(B)=1/3$. Averaging the branch lengths from part (b), the expected length producing the folded $3:1$ class is
$$
B_{3:1}=\tfrac23(4E_4+2E_3+2E_2)+\tfrac13(4E_4+2E_3)
=4E_4+2E_3+\tfrac43E_2.
$$
The factor $2E_2$ in type A includes both the one-descendant and three-descendant root branches. The expected length producing the $2:2$ class is
$$
B_{2:2}=\tfrac23E_3+\tfrac13(E_3+2E_2)
=E_3+\tfrac23E_2.
$$
With a common <mutation rate> per unit branch length, it cancels from the ratio of expected counts in this <folded site frequency spectrum>:
$$
R=\frac{4E_4+2E_3+\tfrac43E_2}{E_3+\tfrac23E_2}.
$$
For the <Yule process>,
$$
\boxed{R_{\rm Yule}=\frac{1+2/3+2/3}{1/3+1/3}=\frac72=3.5}.
$$
For <Kingman's coalescent>,
$$
\boxed{R_{\rm coal}=\frac{2/3+2/3+4/3}{1/3+2/3}=\frac83\approx2.67}.
$$
These are ratios of ensemble expected mutation counts, not the <expected value> of the random ratio in one tree. The stipulated topology weights apply to both processes, and their waiting-time rates depend only on the current number of branches.
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