AMC12 2010 A

We can solve this by breaking the problem down into $2$ cases and adding up the probabilities. Case $1$: Bernardo picks $9$. If Bernardo picks a $9$ then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a $9$ is $\frac{1 \cdot \binom{8}{2}}{\binom{9}{3}} = \frac{\frac{8\cdot7}{2}}{\frac{9\cdot8\cdot7}{3\cdot2\cdot1}}=\frac{1}{3}$. Case $2$: Bernardo does not pick $9$. Since the chance of Bernardo picking $9$ is $\frac{1}{3}$, the probability of not picking $9$ is $\frac{2}{3}$. If Bernardo does not pick $9$, then he can pick any number from $1$ to $8$. Since Bernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger. Ignoring the $9$ for now, the probability that they will pick the same number is the number of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers. We get this probability to be $\frac{3!}{8\cdot{7}\cdot{6}} = \frac{1}{56}$ The probability of Bernardo's number being greater is \[\frac{1-\frac{1}{56}}{2} = \frac{55}{112}\] Factoring the fact that Bernardo could've picked a $9$ but didn't: \[\frac{2}{3}\cdot{\frac{55}{112}} = \frac{55}{168}\] Adding up the two cases we get $\frac{1}{3}+\frac{55}{168} = \boxed{\frac{37}{56}\ \textbf{(B)}}$ We have for case $1$: $\frac{1 \cdot \binom{8}{2}}{\binom{9}{3}}$ since $1$ is the number of ways to pick 9 and $\binom{8}{2}$ is the number of ways to pick the other 2 numbers. $\binom{9}{3}$ means to choose 3 numbers from 9. A common pitfall is saying that the probability of picking the same number is $\frac{8*7*6}{(8*7*6)^2}$. This actually undercounts. Note that picking $3,7,6$ will lead to the same end result as picking $7,3,6$ (order does not matter, since it will be descending no matter what). Thus, we multiply by $3!$ :)