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.
Contents
Problem
Bernardo randomly picks 3 distinct numbers from the set and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set 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?
Solution 1
We can solve this by breaking the problem down into cases and adding up the probabilities.
Case : Bernardo picks .
If Bernardo picks a then it is guaranteed that his number will be larger than Silvia's. The probability that he will pick a is .
Case : Bernardo does not pick .
Since the chance of Bernardo picking is , the probability of not picking is .
If Bernardo does not pick , then he can pick any number from to . 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 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
The probability of Bernardo's number being greater is
Factoring the fact that Bernardo could've picked a but didn't:
Adding up the two cases we get
Note
We have for case : since is the number of ways to pick 9 and is the number of ways to pick the other 2 numbers. means to choose 3 numbers from 9.
~mathboy282
A common pitfall is saying that the probability of picking the same number is . This actually undercounts. Note that picking will lead to the same end result as picking (order does not matter, since it will be descending no matter what). Thus, we multiply by :)
-smartguy888
Solution 2
From Bernardo's set, there are numbers that he can randomly choose. From Silvia's set, there are numbers that she can randomly choose. Since Bernardo and Silvia can choose their numbers independently, there are 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 numbers are all bigger than Silvia's numbers. Here, we have 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 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 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 pairs where his number will be greater than Silvia's.
In total, Bernardo's probability of having a greater number than Silvia's is , which is .
- Leo M.
solution 3
Two cases:
1st: Bernado gets a nine, there are in total ways such Bernado gets a bigger number
2nd: Both of them get numbers from one to eight, because of symmetry, they have equal chance of getting a bigger number. Thus, there are ways
And there are in total ways to pick numbers, the possibility is
~bluesoul
Another idea
Hint: Let Brandon's number be B and Silvia's number be s.
Video Solutions
https://youtu.be/rsURe5Xh-j0?t=590
~IceMatrix
~savannahsolve
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.