2010 AIME II Problems/Problem 13
Contents
[hide]Problem
The cards in a deck are numbered . Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards form a team, and the two persons with higher numbered cards form another team. Let be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards and , and Dylan picks the other of these two cards. The minimum value of for which can be written as . where and are relatively prime positive integers. Find .
Solution
Once the two cards are drawn, there are ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below , which occurs in ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above , which occurs in ways. Thus, Simplifying, we get , so we need . If , then So or , and or , so or . Thus, , and the answer is .
Solution 2
Given that Alex and Dylan hold the cards and , we need to calculate the probability that they end up on the same team. This happens in two scenarios:
1. Both on the Lower Team: This occurs if the other two cards drawn are both greater than . 2. Both on the Higher Team: This occurs if the other two cards drawn are both less than .
The total number of ways to choose the other two cards from the remaining 50 cards is .
The number of favorable outcomes is the sum of: The number of ways to choose 2 cards greater than : The number of ways to choose 2 cards less than :
Thus, the probability is:
Finding the Minimum for
Solving the inequality: leads to:
This quadratic inequality is satisfied when or .
The minimum value of that satisfies occurs at (or symmetrically at ), giving: where and are coprime.
Final Answer
Answer:
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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