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

Revision as of 20:57, 23 July 2017

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

Solution 3: The fastest approach

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 $2-0.5=1.5$ over a total length of $2-0=2$ . Therefore, the probability is 1.5/2=15/20= $\boxed{3/4 \space \text{C}}$

@@ Line 13: / Line 13: @@
 ==Solution 3: The fastest approach==
-Scale down by 2017 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 0.5 to 2 to pick from, on average, which is a length of 2-0.5=1.5. Therefore, the probability is 1.5/2=15/20=<math>\boxed{3/4 \space  \text{C}}</math>
+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 0.5 to 2 to pick from, on average, which is a length of <math>2-0.5=1.5</math> over a total length of <math>2-0=2</math>. Therefore, the probability is 1.5/2=15/20=<math>\boxed{3/4 \space  \text{C}}</math>
 ==See Also==
 {{AMC10 box|year=2017|ab=A|num-b=14|num-a=16}}
 {{MAA Notice}}

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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