Difference between revisions of "2017 AMC 12A Problems/Problem 10"
m |
Lphillips17 (talk | contribs) (→Solution) |
||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
Suppose Laurent's number is in the interval <math> [ 0, 2017 ] </math>. Then, by symmetry, the probability of Laurent's number being greater is <math>\dfrac{1}{2}</math>. Next, suppose Laurent's number is in the interval <math> [ 2017, 4034 ] </math>. Then Laurent's number will be greater with probability <math>1</math>. Since each case is equally likely, the probability of Laurent's number being greater is <math>\dfrac{1 + \frac{1}{2}}{2} = \dfrac{3}{4}</math>, so the answer is C. | Suppose Laurent's number is in the interval <math> [ 0, 2017 ] </math>. Then, by symmetry, the probability of Laurent's number being greater is <math>\dfrac{1}{2}</math>. Next, suppose Laurent's number is in the interval <math> [ 2017, 4034 ] </math>. Then Laurent's number will be greater with probability <math>1</math>. Since each case is equally likely, the probability of Laurent's number being greater is <math>\dfrac{1 + \frac{1}{2}}{2} = \dfrac{3}{4}</math>, so the answer is C. | ||
| + | |||
| + | ==Alternate Solution: Geometric Probability== | ||
| + | Let <math>x</math> be the number chosen randomly by Chloé. Because it is given that the number Chloé choose is 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 Chloé choose is interval <math> [ 0, 2017 ] </math>, <math> 0 \leq y \leq 4034</math>. Since we are looking for when Laurent's number is Chloé'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 the all the possibilities and 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. The simplified quotient of these two areas is the probability Laurent's number is larger than Chloé's, which is <math>\boxed \dfrac{3}{4} = C</math>. | ||
==See Also== | ==See Also== | ||
Revision as of 15:16, 9 February 2017
Problem
Chloé 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
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 probability 
. Since each case is equally likely, the probability of Laurent's number being greater is 
, so the answer is C.
Alternate Solution: Geometric Probability
Let 
be the number chosen randomly by Chloé. Because it is given that the number Chloé choose is interval , 
. Next, let 
be the number chosen randomly by Laurent. Because it is given that the number Chloé choose is interval , 
. Since we are looking for when Laurent's number is Chloé's we write the equation 
. When these three inequalities are graphed the area captured by and represents the all the possibilities and the area captured by , , and represents the possibilities of Laurent winning. The simplified quotient of these two areas is the probability Laurent's number is larger than Chloé's, which is $\boxed \dfrac{3}{4} = C$ (Error compiling LaTeX. Unknown error_msg).
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. 
