Difference between revisions of "2017 AMC 12A Problems/Problem 10"
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<math> \textbf{(A)}\ \dfrac{1}{2} \qquad\textbf{(B)}\ \dfrac{2}{3} \qquad\textbf{(C)}\ \dfrac{3}{4} \qquad\textbf{(D)}\ \dfrac{5}{6} \qquad\textbf{(E)}\ \dfrac{7}{8} </math> | <math> \textbf{(A)}\ \dfrac{1}{2} \qquad\textbf{(B)}\ \dfrac{2}{3} \qquad\textbf{(C)}\ \dfrac{3}{4} \qquad\textbf{(D)}\ \dfrac{5}{6} \qquad\textbf{(E)}\ \dfrac{7}{8} </math> | ||
| + | |||
| + | ==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. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2017|ab=A|num-b=9|num-a=11}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:14, 8 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.
See Also
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
