Difference between revisions of "2017 AMC 10A Problems/Problem 15"
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− | Suppose a point <math>(x,y)</math> lies in the <math>xy</math>-plane. Let <math>x</math> be Chloe's number and <math>y</math> be Laurent's number. Then obviously we want <math>y>x</math>, which basically gives us a region above a line. We know that Chloe's number is in the interval <math>[0,2017]</math> and Laurent's number is in the interval <math>[0,4034]</math>, so we can create a rectangle in the plane, whose length is <math>2017</math> and whose width is <math>4034</math>. 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. \boxed{\textbf{(C)}\ \frac{3}{4}} | + | Suppose a point <math>(x,y)</math> lies in the <math>xy</math>-plane. Let <math>x</math> be Chloe's number and <math>y</math> be Laurent's number. Then obviously we want <math>y>x</math>, which basically gives us a region above a line. We know that Chloe's number is in the interval <math>[0,2017]</math> and Laurent's number is in the interval <math>[0,4034]</math>, so we can create a rectangle in the plane, whose length is <math>2017</math> and whose width is <math>4034</math>. 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. <math>\boxed{\textbf{(C)}\ \frac{3}{4}}</math>. |
==Solution 3== | ==Solution 3== |
Revision as of 16:40, 15 April 2023
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.