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

Revision as of 11:49, 2 July 2019

Problem

Chloé 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 Chloé's number? (Assume they cannot be equal)

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

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 $[2017, 4034]$ . 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 $\frac{1}{2}*1 + \frac{1}{2}*\frac{1}{2} = \boxed{\textbf{(C)}\ \frac{3}{4}}$

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

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 $1.5/2=15/20=\boxed{3/4 \space \text{(C)}}$

@@ Line 6: / Line 6: @@
 ==Solution 1==
 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>[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}*1 + \frac{1}{2}*\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>[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{\textbf{(C)}\ \frac{3}{4}}</math>
 ==Solution 2==

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