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

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

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

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