Difference between revisions of "2018 AMC 12A Problems/Problem 24"
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| − | + | == Problem == | |
Alice, Bob, and Carol play a game in which each of them chooses a real number between 0 and 1. The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between 0 and 1, and Bob announces that he will choose his number uniformly at random from all the numbers between <math>\tfrac{1}{2}</math> and <math>\tfrac{2}{3}.</math> Armed with this information, what number should Carol choose to maximize her chance of winning? | Alice, Bob, and Carol play a game in which each of them chooses a real number between 0 and 1. The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between 0 and 1, and Bob announces that he will choose his number uniformly at random from all the numbers between <math>\tfrac{1}{2}</math> and <math>\tfrac{2}{3}.</math> Armed with this information, what number should Carol choose to maximize her chance of winning? | ||
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</math> | </math> | ||
| − | + | == Solution == | |
Plug in all the answer choices to get <math>\boxed{\textbf{(B)}.}</math> | Plug in all the answer choices to get <math>\boxed{\textbf{(B)}.}</math> | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2018|ab=A|num-b=23|num-a=25}} | ||
| + | {{MAA Notice}} | ||
Revision as of 14:39, 8 February 2018
Problem
Alice, Bob, and Carol play a game in which each of them chooses a real number between 0 and 1. The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between 0 and 1, and Bob announces that he will choose his number uniformly at random from all the numbers between 
and 
Armed with this information, what number should Carol choose to maximize her chance of winning?
Solution
Plug in all the answer choices to get 
See Also
| 2018 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 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. 
