AMC 12 Mock Exam 5 Problem 22

Only the pattern of bill types matters, and all $\binom{10}{5}=252$ patterns are equally likely. Let $h_j$ be the number of $\$5$ bills in the register after $j$ customers; $h_0=1$, a $\$5$ customer adds $1$, a $\$10$ customer subtracts $1$, and $h_{10}=1$. The cashier succeeds iff $h_j\ge0$ for all $j$, i.e., the path never reaches $-1$.

Reflection. For a path from height $1$ to height $1$ that reaches $-1$, reflect the part before the first visit to $-1$ across the level $-1$. This gives a bijection with all paths of $10$ steps from height $-3$ to height $1$. Such a path has $7$ up-steps and $3$ down-steps, so there are $\binom{10}{3}=120$ of them.

So $252-120=132$ patterns succeed, and the probability is $\frac{132}{252}=\boxed{\tfrac{11}{21}}$. (With an empty register the answer would be the Catalan count $\frac{42}{252}=\frac16$.)