2017 AMC 10A Problems/Problem 15
Contents
Problem
Chloe chooses a real number uniformly at random from the interval . Independently, Laurent chooses a real number uniformly at random from the interval . What is the probability that Laurent's number is greater than Chloe's number?
Solution 1
Denote "winning" to mean "picking a greater number". There is a chance that Laurent chooses a number in the interval . 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
~Small grammar mistake corrected by virjoy2001 (missing period), small error corrected by Terribleteeth
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 and Laurent's number is in the interval , so we can create a rectangle in the plane, whose length is and whose width is . Drawing it out and dividing into 4 congruent triangles, we see that Laurent's winning area is 3 triangles and Chloe's is 1 triangle. .
Solution 3
Scale down by to get that Chloe picks from and Laurent picks from . There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of . Therefore, Laurent has a winning range of , where the average value of is . Thus the average winning length is out of a total length of . Therefore, the probability is
Video Solution
A video solution for this can be found here: https://www.youtube.com/watch?v=PQFNwW1XFaQ
~savannahsolver
Video Solution 2
Video Solution
https://youtu.be/IRyWOZQMTV8?t=4163
~ pi_is_3.14
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.