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

Difference between revisions of "2017 AMC 10A Problems/Problem 15"

(→‎Solution 2)
m (→‎Solution)
Line 4: Line 4:
 
<math> \mathrm{(A) \ }\frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{4}\qquad \mathrm{(D) \ } \frac{5}{6}\qquad \mathrm{(E) \ }\frac{7}{8}</math>
 
<math> \mathrm{(A) \ }\frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{4}\qquad \mathrm{(D) \ } \frac{5}{6}\qquad \mathrm{(E) \ }\frac{7}{8}</math>
  
−
==Solution==
+
==Solution1==
 
Denote "winning" to mean "picking a greater number".
 
Denote "winning" to mean "picking a greater number".
−
There is a <math>\frac{1}{2}</math> chance that Laurent chooses a number in the interval <math>(2017, 4032]</math>. In this case, Chloé cannot possibly win, since the maximum number she can pick is <math>2017</math>. Otherwise, if Laurent picks a number in the interval <math>[0, 2017]</math>, with probability <math>\frac{1}{2}</math>, then the two people are symmetric, and each has a <math>\frac{1}{2}</math> chance of winning. Then, the total probability is <math>\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{\frac{3}{4} (C)}</math>
+
There is a <math>\frac{1}{2}</math> chance that Laurent chooses a number in the interval <math>(2017, 4032]</math>. In this case, Chloé cannot possibly win, since the maximum number she can pick is <math>2017</math>. Otherwise, if Laurent picks a number in the interval <math>[0, 2017]</math>, with probability <math>\frac{1}{2}</math>, then the two people are symmetric, and each has a <math>\frac{1}{2}</math> chance of winning. Then, the total probability is <math>\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{(C) \frac{3}{4}}</math>
−
 
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 22:39, 8 February 2017

Problem

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

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

Solution1

Denote "winning" to mean "picking a greater number". There is a $\frac{1}{2}$ chance that Laurent chooses a number in the interval $(2017, 4032]$. 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 $\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{(C) \frac{3}{4}}$

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, 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 $1$. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line $y>x$, which is $\frac{2017 \cdot 2017}{2}$. Instead of bashing this out we know that the rectangle has area $2017 \cdot 4034$. So the probability that Laurent has a smaller number is $\frac{2017 \cdot 2017}{2 \cdot 2017 \cdot 4034}$. Simplifying the expression yields $\frac{1}{4}$ and so $1-\frac{1}{4}= \boxed{\frac{3}{4} (C)}$.

~AoPS12142015

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=83232"