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2010 AMC 10A Problems/Problem 18

Revision as of 23:53, 19 November 2019 by Sanspapyrus683 (talk | contribs) (→‎Solution)

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 $\dfrac{3}{9} = \frac{1}{3}$.


Case $2$: Bernardo does not pick $9$. Since the chance of Bernardo picking $9$ is $\frac{1}{9}$, the probability of not picking $9$ is $\frac{8}{9}$.

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

See also

2010 AMC 10A (Problems • Answer Key • Resources)
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

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