2010 AMC 12A Problems/Problem 16

The following problem is from both the 2010 AMC 12A #16 and 2010 AMC 10A #18, so both problems redirect to this page.

Problem

Bernardo randomly picks 3 distinct numbers from the set $\{1,2,3,4,5,6,7,8,9\}$ and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set $\{1,2,3,4,5,6,7,8\}$ and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?

$\textbf{(A)}\ \frac{47}{72} \qquad \textbf{(B)}\ \frac{37}{56} \qquad \textbf{(C)}\ \frac{2}{3} \qquad \textbf{(D)}\ \frac{49}{72} \qquad \textbf{(E)}\ \frac{39}{56}$

Solution 1

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 $\frac{3!}{8\cdot{7}\cdot{6}} = \frac{1}{56}$

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 $\frac{1}{3}+\frac{55}{168} = \boxed{\frac{37}{56}\ \textbf{(B)}}$

~Edits by mathboy282

Note

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 rest 2 numbers. $\binom{9}{3}$ is just from 9 numbers pick 3.

~nerdddd

Solution 2

From Bernardo's set, there are $\binom{9}{3} = 84$ numbers that he can randomly choose. From Silvia's set, there are $\binom{8}{3} = 56$ numbers that she can randomly choose. Since Bernardo and Silvia can choose their numbers independently, there are $84 \cdot 56$ pairs of numbers that you can compare. For example, if Bernardo chooses 321 and Silvia chooses 543, that is one pair. We can sort Bernardo's numbers from the greatest to the smallest. We can do the same for Silvia's numbers. So, Bernardo's greatest $84 - 56 = 28$ numbers are all bigger than Silvia's numbers. Here, we have $28 \cdot 56$ pairs satisfying that Bernardo's number will be greater than Silvia's. If Bernardo chooses the 29th greatest number, which is the same as Silvia's greatest number, hence there will be $56 - 1 = 55$ pairs satisfying that Bernardo's 29th greatest number will be greater than Silvia's. Similarly, if Bernardo chooses the 30th greatest number, there will be $55 - 1 = 54$ pairs satisfying that Bernardo's 30th greatest number will be greater than Silvia's. This pattern continues. So if Bernardo chooses the 29th greatest number or below, he will have $55 + 54 + 53 + \cdots + 1$ pairs where his number will be greater than Silvia's.

In total, Bernardo's probability of having a greater number than Silvia's is $\frac{28 \cdot 56 + (55 + 54 + 53 + \cdots + 1)}{84 \cdot 56} = \frac{37}{56}$, which is $\textbf{(B)}$.


- Leo M.

Video Solution by the Beauty of Math

https://youtu.be/rsURe5Xh-j0?t=590

See also

2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2010 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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