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

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==Problem==
 
==Problem==
Chloé chooses a real number uniformly at random from the interval <math>[0, 2017]</math>. Independently, Laurent cooses a real number uniformly at random from the interval <math>[0, 4034]</math>. What is the probability that Laurent's number is greater than Chloé's number?
 
  
<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>
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Chloe chooses a real number uniformly at random from the interval <math>[0, 2017]</math>. Independently, Laurent chooses a real number uniformly at random from the interval <math>[0, 4034]</math>. What is the probability that Laurent's number is greater than Chloe's number?
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<math> \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}</math>
  
 
==Solution 1==
 
==Solution 1==
 
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, 4034]</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{\bold{(C)}\ \frac{3}{4}}</math>
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There is a <math>\frac{1}{2}</math> chance that Laurent chooses a number in the interval <math>[2018, 4034]</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}\times1 + \frac{1}{2}\times\frac{1}{2} = \boxed{\textbf{(C)}\ \frac{3}{4}}.</math>
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~Small grammar mistake corrected by virjoy2001 (missing period), small error corrected by Terribleteeth
  
 
==Solution 2==
 
==Solution 2==
 
We can use geometric probability to solve this.
 
We can use geometric probability to solve this.
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, 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 <math>1</math>. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line <math>y>x</math>, which is <math>\frac{2017 \cdot 2017}{2}</math>. Instead of bashing this out we know that the rectangle has area <math>2017 \cdot 4034</math>. So the probability that Laurent has a smaller number is <math>\frac{2017 \cdot 2017}{2 \cdot 2017 \cdot 4034}</math>. Simplifying the expression yields <math>\frac{1}{4}</math> and so <math>1-\frac{1}{4}= \boxed{\bold{(C)}\ \frac{3}{4}}</math>.  
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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, 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 <math>1</math>. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line <math>y>x</math>, which is <math>\frac{2017 \cdot 2017}{2}</math>. Instead of bashing this out we know that the rectangle has area <math>2017 \cdot 4034</math>. So the probability that Laurent has a smaller number is <math>\frac{2017 \cdot 2017}{2 \cdot 2017 \cdot 4034}</math>. Simplifying the expression yields <math>\frac{1}{4}</math> and so <math>1-\frac{1}{4}= \boxed{\textbf{(C)}\ \frac{3}{4}}</math>.
  
~AoPS12142015
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==Solution 3==
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Scale down by <math>2017</math> to get that Chloe picks from <math>[0,1]</math> and Laurent picks from <math>[0,2]</math>. There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of <math>0.5</math>. Therefore, Laurent has a range of <math>0.5</math> to <math>2</math> to pick from, on average, which is a length of <math>2-0.5=1.5</math> out of a total length of <math>2-0=2</math>. Therefore, the probability is <math>1.5/2=15/20=\boxed{\frac{3}{4} \space \text{(C)}}.</math>
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==Video Solution==
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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==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}}
{{AMC12 box|year=2017|ab=A|num-b=9|num-a=11}}
 
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Probability Problems]]

Revision as of 01:31, 7 December 2021

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 $[2018, 4034]$. 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, 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 $1-\frac{1}{4}= \boxed{\textbf{(C)}\ \frac{3}{4}}$.

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 (ProblemsAnswer KeyResources)
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

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