Difference between revisions of "2017 AMC 10A Problems/Problem 15"
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==Problem== | ==Problem== | ||
| − | Chloé chooses a real number uniformly at random from the interval <math>[0, 2017]</math>. Independently, Laurent | + | Chloé chooses a real number uniformly at random from the interval <math>[0, 2017]</math>. Independently, Laurent chooses a real number uniformly at random from the interval <math>[0, 4034]</math>. What is the probability that Laurent's number is greater than Chloé's number? |
<math> \mathrm{\textbf{(A)} \ }\frac{1}{2}\qquad \mathrm{\textbf{(B)} \ } \frac{2}{3}\qquad \mathrm{\textbf{(C)} \ } \frac{3}{4}\qquad \mathrm{\textbf{(D)} \ } \frac{5}{6}\qquad \mathrm{\textbf{(E)} \ }\frac{7}{8}</math> | <math> \mathrm{\textbf{(A)} \ }\frac{1}{2}\qquad \mathrm{\textbf{(B)} \ } \frac{2}{3}\qquad \mathrm{\textbf{(C)} \ } \frac{3}{4}\qquad \mathrm{\textbf{(D)} \ } \frac{5}{6}\qquad \mathrm{\textbf{(E)} \ }\frac{7}{8}</math> | ||
Revision as of 20:33, 15 February 2017
Contents
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 chooses 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 1
Denote "winning" to mean "picking a greater number".
There is a 
chance that Laurent chooses a number in the interval ![$[2017, 4034]$](http://latex.artofproblemsolving.com/f/d/c/fdce99d940d567c0241a5dca0a21a8a77f631351.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 
Solution 2
We can use geometric probability to solve this.
Suppose a point 
lies in the 
-plane. Let 
be Chloe's number and 
be Laurent's number. Then obviously we want 
, which basically gives us a region above a line. We know that Chloe's number is in the interval ![$[0,2017]$](http://latex.artofproblemsolving.com/2/9/2/2923eaa5fb158c9fb73b5c3184f250517c9325da.png)
and Laurent's number is in the interval ![$[0,4034]$](http://latex.artofproblemsolving.com/e/4/c/e4ce3d7d085bbe0844278017306c1118e35e3d6c.png)
, so we can create a rectangle in the plane, whose length is and whose width is 
. Drawing it out, we see that it is easier to find the probability that Chloe's number is greater than Laurent's number and subtract this probability from 
. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line , which is 
. Instead of bashing this out we know that the rectangle has area 
. So the probability that Laurent has a smaller number is 
. Simplifying the expression yields 
and so 
.
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 | ||
| 2017 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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. 
