= Solution
After three multiplicative steps, $k$ upward moves and $3-k$ downward moves give
$$
S_3(k)=100(1.1)^k(0.9)^{3-k}.
$$
For $k=0,1,2,3$, respectively, the prices are $\boxed{72.9,\ 89.1,\ 108.9,\ 133.1}$. Since the steps are independent, the endpoint <probabilities> are binomial:
$$
\boxed{\Pr(S_3=S_3(k))=\binom3k p^k(1-p)^{3-k}.}
$$
In the displayed ascending-price order they are $(1-p)^3$, $3p(1-p)^2$, $3p^2(1-p)$, and $p^3$. Different paths with the same count of up moves recombine because multiplication by $1.1$ and $0.9$ commutes.
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