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

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==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{(C) \frac{3}{4}}</math>
+
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>
  
 
==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{(C) \frac{3}{4}}</math>.  
+
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>.  
  
 
~AoPS12142015
 
~AoPS12142015

Revision as of 14:17, 9 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?

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

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 $\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{\bold{(C)}\ \frac{3}{4}}$

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 $1-\frac{1}{4}= \boxed{\bold{(C)}\ \frac{3}{4}}$.

~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
2017 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 12 Problems and Solutions

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