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

(Problem)
m (Solution 1)
 
(2 intermediate revisions by 2 users not shown)
Line 7: Line 7:
 
==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}\times1 + \frac{1}{2}\times\frac{1}{2} = \boxed{\textbf{(C)}\ \frac{3}{4}}.</math>
+
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>
  
~Small grammar mistake corrected by virjoy2001 (missing period)
+
~Small grammar mistake corrected by virjoy2001 (missing period), small error corrected by Terribleteeth
  
 
==Solution 2==
 
==Solution 2==
Line 27: Line 27:
 
==Video Solution 2==
 
==Video Solution 2==
 
https://youtu.be/s4vnGlwwHHw
 
https://youtu.be/s4vnGlwwHHw
 +
 +
== Video Solution ==
 +
https://youtu.be/IRyWOZQMTV8?t=4163
 +
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 00:36, 27 January 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS