Difference between revisions of "2017 AMC 10A Problems/Problem 15"
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==Solution 2== | ==Solution 2== | ||
We can use geometric probability to solve this. | We can use geometric probability to solve this. | ||
− | Suppose a point <math>(x,y)</math> lies in the <math>xy</math>-plane. Let <math>x</math> be Chloe's number and <math>y</math> be Laurent's number. Then obviously we want <math>y>x</math>, which basically gives us a region above a line. We know that Chloe's number is in the interval <math>[0,2017]</math> and Laurent's number is in the interval <math>[0,4034]</math>, so we can create a rectangle in the plane, whose length is <math>2017</math> and whose width is <math>4034</math>. 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 <math>1</math>. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line <math>y>x</math>, which is <math>\frac{2017 \cdot 2017}{2}</math>. Instead of bashing this out we know that the rectangle has area <math>2017 \cdot 4034</math>. So the probability that Laurent has a smaller number is <math>\frac{2017 \cdot 2017}{2 \cdot 2017 \cdot 4034}</math>. Simplifying the expression yields <math>\frac{1}{4}</math> and so <math>1-\frac{1}{4}= \boxed{\textbf{(C)}\ \frac{3}{4}}</math>. | + | Suppose a point <math>(x,y)</math> lies in the <math>xy</math>-plane. Let <math>x</math> be Chloe's number and <math>y</math> be Laurent's number. Then obviously we want <math>y>x</math>, which basically gives us a region above a line. We know that Chloe's number is in the interval <math>[0,2017]</math> and Laurent's number is in the interval <math>[0,4034]</math>, so we can create a rectangle in the plane, whose length is <math>2017</math> and whose width is <math>4034</math>. 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 <math>1</math>. The probability that Chloe's number is larger than Laurent's number is simply the area of the region under the line <math>y>x</math>, which is <math>\frac{2017 \cdot 2017}{2}</math>. Instead of bashing this out we know that the rectangle has area <math>2017 \cdot 4034</math>. So the probability that Laurent has a smaller number is <math>\frac{2017 \cdot 2017}{2 \cdot 2017 \cdot 4034}</math>. Simplifying the expression yields <math>\frac{1}{4}</math> and so <math>1-\frac{1}{4}= \boxed{\textbf{(C)}\ \frac{3}{4}}</math>. |
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==See Also== | ==See Also== |
Revision as of 23:13, 11 February 2017
Contents
Problem
Chloé chooses a real number uniformly at random from the interval . Independently, Laurent cooses a real number uniformly at random from the interval
. What is the probability that Laurent's number is greater than Chloé'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
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
.
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 |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 12 Problems and Solutions |
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