Difference between revisions of "2010 AIME II Problems/Problem 13"
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== Problem == | == Problem == | ||
− | The <math>52</math> cards in a deck are numbered <math>1, 2, \cdots, 52</math>. 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 | + | The <math>52</math> cards in a deck are numbered <math>1, 2, \cdots, 52</math>. 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 <math>p(a)</math> be the [[probability]] that Alex and Dylan are on the same team, given that Alex picks one of the cards <math>a</math> and <math>a+9</math>, and Dylan picks the other of these two cards. The minimum value of <math>p(a)</math> for which <math>p(a)\ge\frac{1}{2}</math> can be written as <math>\frac{m}{n}</math>. where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
== Solution == | == Solution == |
Latest revision as of 10:15, 7 January 2021
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 .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.