Difference between revisions of "2017 AMC 12A Problems/Problem 10"
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==Solution 2 (Geometric Probability)== | ==Solution 2 (Geometric Probability)== | ||
Let <math>x</math> be the number chosen randomly by Chloe. Because it is given that the number Chloe chooses is in the interval <math> [ 0, 2017 ] </math>, <math> 0 \leq x \leq 2017</math>. Next, let <math>y</math> be the number chosen randomly by Laurent. Because it is given that the number Laurent chooses is in the interval <math> [ 0, 4034 ] </math>, <math> 0 \leq y \leq 4034</math>. Since we are looking for when Laurent's number is greater than Chloe's we write the equation <math>y > x</math>. When these three inequalities are graphed the area captured by <math> 0 \leq x \leq 2017</math> and <math> 0 \leq y \leq 4034</math> represents all the possibilities, forming a rectangle <math>2017</math> in width and <math>4034</math> in height. Thus making its area <math>4034 * 2017</math>. The area captured by <math> 0 \leq x \leq 2017</math>, <math> 0 \leq y \leq 4034</math>, and <math>y > x</math> represents the possibilities of Laurent winning, forming a trapezoid with a height <math>2017</math> in length and bases <math>4034</math> and <math>2017</math> length, thus making an area <math>2017 *\frac{4034+2017}{2}</math>. The simplified quotient of these two areas is the probability Laurent's number is larger than Chloe's, which is <math> \boxed {C=\frac{3}{4}}</math>. | Let <math>x</math> be the number chosen randomly by Chloe. Because it is given that the number Chloe chooses is in the interval <math> [ 0, 2017 ] </math>, <math> 0 \leq x \leq 2017</math>. Next, let <math>y</math> be the number chosen randomly by Laurent. Because it is given that the number Laurent chooses is in the interval <math> [ 0, 4034 ] </math>, <math> 0 \leq y \leq 4034</math>. Since we are looking for when Laurent's number is greater than Chloe's we write the equation <math>y > x</math>. When these three inequalities are graphed the area captured by <math> 0 \leq x \leq 2017</math> and <math> 0 \leq y \leq 4034</math> represents all the possibilities, forming a rectangle <math>2017</math> in width and <math>4034</math> in height. Thus making its area <math>4034 * 2017</math>. The area captured by <math> 0 \leq x \leq 2017</math>, <math> 0 \leq y \leq 4034</math>, and <math>y > x</math> represents the possibilities of Laurent winning, forming a trapezoid with a height <math>2017</math> in length and bases <math>4034</math> and <math>2017</math> length, thus making an area <math>2017 *\frac{4034+2017}{2}</math>. The simplified quotient of these two areas is the probability Laurent's number is larger than Chloe's, which is <math> \boxed {C=\frac{3}{4}}</math>. | ||
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| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/LwtoLiBwO-E?t=79 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
Revision as of 04:37, 4 November 2022
Contents
Problem
Chloe chooses a real number uniformly at random from the interval ![$[ 0,2017 ]$](http://latex.artofproblemsolving.com/c/2/d/c2de3743d64138d4bd7f371fb3de265718621d07.png)
. Independently, Laurent chooses a real number uniformly at random from the interval ![$[ 0 , 4034 ]$](http://latex.artofproblemsolving.com/a/9/1/a917fb72d07cd62fd999921ebed24aebddad250d.png)
. What is the probability that Laurent's number is greater than Chloe's number?
Solution 1
Suppose Laurent's number is in the interval ![$[ 0, 2017 ]$](http://latex.artofproblemsolving.com/7/6/a/76a259806417a270b1859d1dd2db61fb0090d670.png)
. Then, by symmetry, the probability of Laurent's number being greater is 
. Next, suppose Laurent's number is in the interval ![$[ 2017, 4034 ]$](http://latex.artofproblemsolving.com/2/2/e/22eb67bd191cf44619936c4b6abe568adb550602.png)
. Then Laurent's number will be greater with a probability of 
. Since each case is equally likely, the probability of Laurent's number being greater is 
, so the answer is 
.
Solution 2 (Geometric Probability)
Let 
be the number chosen randomly by Chloe. Because it is given that the number Chloe chooses is in the interval , 
. Next, let 
be the number chosen randomly by Laurent. Because it is given that the number Laurent chooses is in the interval ![$[ 0, 4034 ]$](http://latex.artofproblemsolving.com/2/d/a/2da5b6a2bc5003d0fb9c7e7b4f5b52e9291b50b0.png)
, 
. Since we are looking for when Laurent's number is greater than Chloe's we write the equation 
. When these three inequalities are graphed the area captured by and represents all the possibilities, forming a rectangle 
in width and 
in height. Thus making its area 
. The area captured by , , and represents the possibilities of Laurent winning, forming a trapezoid with a height in length and bases and length, thus making an area 
. The simplified quotient of these two areas is the probability Laurent's number is larger than Chloe's, which is 
.
Video Solution by OmegaLearn
https://youtu.be/LwtoLiBwO-E?t=79
~ pi_is_3.14
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
