2010 AMC 12A Problem 16

There are two cases: Bernardo picks a $9$ or he doesn’t. Case $1:$ Bernardo picks a $9$ Since a number is fixed, there are $\binom{8}{2} = 28$ ways to choose the other two numbers. There are a total of $\binom{9}{3} = 84$ ways to pick all three numbers. The probability is then $$\dfrac{28}{84} = \dfrac{1}{3}.$$ Note that if Bernardo picks a $9,$ he automatically has a greater number than Silvia. This means that Bernardo always wins in this case. Case $2:$ Bernardo doesn’t pick a $9$ There is a $1 - \frac{1}{3} = \frac{2}{3}$ chance of this happening. Since both people are choosing from the same numbers, they have an equal chance of winning. We still need to find the probability that the numbers are the same. There is a $$\dfrac{1}{\binom{8}{3}} = \dfrac{1}{56}$$ chance that Silvia chooses the same numbers as Bernardo. The probability that Bernardo gets a higher number is then $$\dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}.$$ The total probability of Bernardo getting a higher number is then $$\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}.$$ Thus, B is the correct answer.