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}\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>[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> |
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+ | ~Small grammar mistake corrected by virjoy2001 (missing period) | ||
==Solution 2== | ==Solution 2== |
Revision as of 21:31, 30 November 2020
Contents
Problem
Chloe chooses a real number uniformly at random from the interval . Independently, Laurent chooses a real number uniformly at random from the interval . What is the probability that Laurent's number is greater than Chloe's number?
Solution 1
Denote "winning" to mean "picking a greater number". There is a chance that Laurent chooses a number in the interval . In this case, Chloé cannot possibly win, since the maximum number she can pick is . Otherwise, if Laurent picks a number in the interval , with probability , then the two people are symmetric, and each has a chance of winning. Then, the total probability is
~Small grammar mistake corrected by virjoy2001 (missing period)
Solution 2
We can use geometric probability to solve this. Suppose a point lies in the -plane. Let be Chloe's number and be Laurent's number. Then obviously we want , which basically gives us a region above a line. We know that Chloe's number is in the interval and Laurent's number is in the interval , so we can create a rectangle in the plane, whose length is and whose width is . 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 . The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line , which is . Instead of bashing this out we know that the rectangle has area . So the probability that Laurent has a smaller number is . Simplifying the expression yields and so .
Solution 3
Scale down by to get that Chloe picks from and Laurent picks from . There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of . Therefore, Laurent has a range of to to pick from, on average, which is a length of out of a total length of . Therefore, the probability is
Video Solution
A video solution for this can be found here: https://www.youtube.com/watch?v=PQFNwW1XFaQ
Video Solution 2
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 |
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