= Solution
Superpose the two independent <Poisson processes>. The combined process has rate $9$, and each arrival is independently a Number 1 bus with probability $1/9$ and a Number 8 bus with probability $8/9$. Therefore the number of Number 8 buses before the first Number 1 has the <geometric distribution> on $\{0,1,2,\ldots\}$, and
$$
\mathbb P(\text{exactly three Number 8 buses})
=\left(\frac89\right)^3\frac19
=\boxed{\frac{512}{6561}}.
$$
This is <number of one Poisson type before another> with rates $1$ and $8$.
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