jmc

We will calculate the probability that Sam can buy his favorite toy just using his 8 quarters and then subtract that from 1 to get the probability that we're looking for. There are a total of $8!$ orders in which the 8 toys can be dispensed. If his favorite toy is the first one that the machine selects, then he can obviously buy it just using his quarters, and then there are $7!$ order in which the other toys can be dispensed, which gives us 7! orderings of the toys that let him buy his favorite just using the quarters. If the first toy is the one that only costs 25 cents, then he will be able to buy his favorite toy just using his remaining quarters if and only if it is the second one dispensed. If those are the first two toys dispensed, there are $6!$ ways to order the other toys which means that we have another $6!$ orderings of the toys that allow him to buy his favorite toy without getting change for the 10 dollar bill. If the first toy costs more than 25 cents, or if two toys are dispensed before his favorite one, then he won't have enough quarters to buy his favorite one without getting change for his ten dollar bill. So out of the $8!$ orders in which the 8 toys can be dispensed, there are $7!+6!$ ways that allow him to buy his favorite toy just using his quarters for a probability of $\frac{7!+6!}{8!}=\frac{6!}{6!}\cdot \frac{7+1}{8\cdot 7}=\frac{1}{7}$. But that is the probability that what we want $\text{doesn’t}$ happen, so we need to subtract it from 1 to get our final probability of $1-\frac{1}{7}=\boxed{\frac{6}{7}}$.