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| − | == Problem ==
| + | #REDIRECT [[2010_AMC_12A_Problems/Problem_16]] |
| − | Bernardo randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8,9\}</math> and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set <math>\{1,2,3,4,5,6,7,8\}</math> 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?
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| − | <math>\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}</math>
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| − | == Solution ==
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| − | We can solve this by breaking the problem down into <math>2</math> cases and adding up the probabilities.
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| − | Case <math>1</math>: Bernardo picks <math>9</math>.
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| − | If Bernardo picks a <math>9</math> then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a <math>9</math> is <math>\dfrac{1 \cdot \binom{8}{2}}{\binom{9}{3}} = \frac{1}{3}</math>.
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| − | Case <math>2</math>: Bernardo does not pick <math>9</math>.
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| − | Since the chance of Bernardo picking <math>9</math> is <math>\frac{1}{3}</math>, the probability of not picking <math>9</math> is <math>\frac{2}{3}</math>.
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| − | If Bernardo does not pick 9, then he can pick any number from <math>1</math> to <math>8</math>. 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.
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| − | Ignoring the <math>9</math> 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.
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| − | We get this probability to be <math>\frac{3!}{8\cdot{7}\cdot{6}} = \frac{1}{56}</math>
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| − | Probability of Bernardo's number being greater is
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| − | <cmath>\frac{1-\frac{1}{56}}{2} = \frac{55}{112}</cmath>
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| − | Factoring the fact that Bernardo could've picked a <math>9</math> but didn't:
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| − | <cmath>\frac{2}{3}\cdot{\frac{55}{112}} = \frac{55}{168}</cmath>
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| − | Adding up the two cases we get <math>\frac{1}{3}+\frac{55}{168} = \boxed{\frac{37}{56}\ \textbf{(B)}}</math>
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| − | == Solution 2 ==
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| − | There are <math>\binom{9}{3}\cdot\binom{8}{3}=84\cdot{56}</math> total pairings of numbers between the two.
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| − | Now, we shall inspect the two cases when Bernardo chooses a <math>9</math> or doesn't.
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| − | Case 1: Bernardo chooses a <math>9</math>.
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| − | There are <math>\binom{8}{2}\cdot\binom{8}{3}=28\cdot{56}</math> total pairings for this.
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| − | Case 2: Bernardo does not choose a <math>9</math>.
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| − | There are <math>\frac{\binom{8}{3}\cdot\binom{8}{3}-\binom{8}{3}}{2}</math> ways for this case.
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| − | Now, add everything up and get <math>\boxed{B}</math>
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| − | == See also ==
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| − | {{AMC10 box|year=2010|num-b=17|num-a=19|ab=A}}
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| − | [[Category:Introductory Combinatorics Problems]] | |
| − | {{MAA Notice}}
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