2017 AMC 10A Problems/Problem 15

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?

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

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 winning range of $[X, 2]$ , where the average value of $X$ is $0.5$ . Thus the average winning length is $2-0.5=1.5$ out of a total length of $2-0=2$ . Therefore, the probability is $1.5/2=15/20=\boxed{\frac{3}{4} \space \text{(C)}}.$

Solution 4

In total there are $2018\times4034$ ways in which Laurene and Chloe can choose numbers (as same number cannot be chosen by both). If Chloe chooses 2017, then Lauren has 2017 ways to win, if Chloe chooses 2016, Lauren has 2018 ways to win and so on until if Chloe chooses 0, Lauren has 4034 ways to win. Thus the answer is:

$\frac{\text{number of ways lauren can win}}{\text{number of combinations}}=\frac{2017+2018+2019...+4034}{2018\times4034}$

Using arithmetic series formula:

$\frac{\frac{1}{2}(2018)(2017+4034)}{2018\times4034}=\frac{3}{4}$

$\fbox{C}$ is the correct answer.

~ Lion08

Solution 5

If Chloé chooses $2017$ , she has a $\frac{1}{2}$ chance of choosing the greater number. If she chooses $2016$ , she has a slightly lower chance, and so on. If she chooses $0$ , then she has a chance of $0$ . Her average chance is $\frac{0+\frac{1}{2}}{2} = \frac{1}{4}$ . So, Laurent's chance of choosing the greater number is $1-\frac{1}{4}=$ $\boxed{\textbf{(C)}\ \frac{3}{4}}$ .

~ stellarferret24

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

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

Page

Toolbox

Search

2017 AMC 10A Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Video Solution

Video Solution 2

Video Solution

See Also