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2010 AMC 12A Problems/Problem 16

Revision as of 19:45, 13 December 2020 by Mathboy282 (talk | contribs) (→‎Solution)
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

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.

~mathboy282

Video Solution by the Beauty of Math

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

See also

2010 AMC 12A (Problems • Answer Key • Resources)
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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