= Solution
Fix any starting village and let $A$ be the pickup distance, $B$ the subsequent passenger distance. From each vertex, the four possible pickup distances are $0,5,10,15$, each with <probability> $1/4$. From each pickup village, the three other possible destinations have distances $5,10,15$, each with <probability> $1/3$.
Sharing the pickup village does not by itself give <independence>. Here it does hold because the <conditional distribution> of $B$ is the same for every pickup village: for all allowed $a,b$,
$$
P(A=a,B=b)=\sum_{u:\,d(\text{start},u)=a}P(U=u)P(B=b\mid U=u)
=P(A=a)\frac13=P(A=a)P(B=b).
$$
Thus $D=A+B$ has the <taxi distance convolution on a rectangular road cycle>. Counting the twelve equally likely pairs gives
$$
\boxed{\begin{array}{c|rrrrrr}
d\text{ (miles)}&5&10&15&20&25&30\\\hline
P(D=d)&1/12&2/12&3/12&3/12&2/12&1/12
\end{array}}
$$
No other distances are possible. The <probability distribution> is the same for every starting village, so no assumption about the previous customer's destination is needed.
The separate moments are $\mathbb EA=15/2$, $\mathbb EB=10$, $\operatorname{Var}A=125/4$ and $\operatorname{Var}B=50/3$. <Independence> therefore gives
$$
\boxed{\mathbb ED=\frac{35}{2}\text{ miles},\qquad
\operatorname{Var}D=\frac{125}{4}+\frac{50}{3}=\frac{575}{12}\text{ miles}^2.}
$$
As a check from the displayed <probability mass function>, $\mathbb ED^2=2125/6$ and subtracting $(35/2)^2$ gives the same <variance>.
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