Difference between revisions of "2017 AMC 10A Problems/Problem 15"
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Denote "winning" to mean "picking a greater number". | Denote "winning" to mean "picking a greater number". | ||
There is a <math>\frac{1}{2}</math> chance that Laurent chooses a number in the interval <math>(2017, 4032]</math>. In this case, Chloé cannot possibly win, since the maximum number she can pick is <math>2017</math>. Otherwise, if Laurent picks a number in the interval <math>[0, 2017]</math>, with probability <math>\frac{1}{2}</math>, then the two people are symmetric, and each has a <math>\frac{1}{2}</math> chance of winning. Then, the total probability is <math>\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{\frac{3}{4} (C)}</math> | There is a <math>\frac{1}{2}</math> chance that Laurent chooses a number in the interval <math>(2017, 4032]</math>. In this case, Chloé cannot possibly win, since the maximum number she can pick is <math>2017</math>. Otherwise, if Laurent picks a number in the interval <math>[0, 2017]</math>, with probability <math>\frac{1}{2}</math>, then the two people are symmetric, and each has a <math>\frac{1}{2}</math> chance of winning. Then, the total probability is <math>\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{\frac{3}{4} (C)}</math> | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:40, 8 February 2017
Problem
Chloé chooses a real number uniformly at random from the interval ![$[0, 2017]$](http://latex.artofproblemsolving.com/1/0/e/10eba1aa7b7e8ac41f2ca92850bdca6ca8d5c4b7.png)
. Independently, Laurent cooses a real number uniformly at random from the interval ![$[0, 4034]$](http://latex.artofproblemsolving.com/4/0/9/409b3cbc7fc05da96f487c86b345dbae7ee98d6e.png)
. What is the probability that Laurent's number is greater than Chloé's number?
Solution
Denote "winning" to mean "picking a greater number".
There is a 
chance that Laurent chooses a number in the interval ![$(2017, 4032]$](http://latex.artofproblemsolving.com/7/8/7/787ab75dd2f832b49eec020275f6e890ff5b90e1.png)
. In this case, Chloé cannot possibly win, since the maximum number she can pick is 
. Otherwise, if Laurent picks a number in the interval , with probability , then the two people are symmetric, and each has a chance of winning. Then, the total probability is 
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
