Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2017 AMC 10A Problems/Problem 15

Revision as of 19:00, 1 April 2023 by Oinava (talk | contribs) (→‎Solution 2)

Problem

Chloe chooses a real number uniformly at random from the interval $[0, 2017]$. Independently, Laurent chooses a real number uniformly at random from the interval $[0, 4034]$. What is the probability that Laurent's number is greater than Chloe's number?

$\mathrm{\textbf{(A)} \ }\frac{1}{2}\qquad \mathrm{\textbf{(B)} \ } \frac{2}{3}\qquad \mathrm{\textbf{(C)} \ } \frac{3}{4}\qquad \mathrm{\textbf{(D)} \ } \frac{5}{6}\qquad \mathrm{\textbf{(E)} \ }\frac{7}{8}$

Solution 1

Denote "winning" to mean "picking a greater number". There is a $\frac{1}{2}$ chance that Laurent chooses a number in the interval $[2017, 4034]$. In this case, Chloé cannot possibly win, since the maximum number she can pick is $2017$. Otherwise, if Laurent picks a number in the interval $[0, 2017]$, with probability $\frac{1}{2}$, then the two people are symmetric, and each has a $\frac{1}{2}$ chance of winning. Then, the total probability is $\frac{1}{2}\times1 + \frac{1}{2}\times\frac{1}{2} = \boxed{\textbf{(C)}\ \frac{3}{4}}.$

~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 $(x,y)$ lies in the $xy$-plane. Let $x$ be Chloe's number and $y$ be Laurent's number. Then obviously we want $y>x$, which basically gives us a region above a line. We know that Chloe's number is in the interval $[0,2017]$ and Laurent's number is in the interval $[0,4034]$, so we can create a rectangle in the plane, whose length is $2017$ and whose width is $4034$. 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}}$.

Solution 3

Scale down by $2017$ to get that Chloe picks from $[0,1]$ and Laurent picks from $[0,2]$. There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of $0.5$. Therefore, Laurent has a range of $0.5$ to $2$ to pick from, on average, which is a length of $2-0.5=1.5$ out of a total length of $2-0=2$. Therefore, the probability is $1.5/2=15/20=\boxed{\frac{3}{4} \space \text{(C)}}.$

Video Solution

A video solution for this can be found here: https://www.youtube.com/watch?v=PQFNwW1XFaQ

https://youtu.be/NB4KXQiqgi0

~savannahsolver

Video Solution 2

https://youtu.be/s4vnGlwwHHw

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_15&oldid=191690"