Difference between revisions of "2003 AMC 12A Problems/Problem 22"
(→Solution) |
(→Solution) |
||
| Line 13: | Line 13: | ||
== Solution == | == Solution == | ||
| − | If <math>A</math> and <math>B</math> meet, their paths connect <math>(0,0)</math> and <math>(5,7).</math> There are <math>\binom{12}{5}=792</math> such paths. Since the path is <math>12</math> units long, they must meet after each travels <math>6</math> units, so the probability is <math>\frac{792}{2^{6}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}</math> | + | If <math>A</math> and <math>B</math> meet, their paths connect <math>(0,0)</math> and <math>(5,7).</math> There are <math>\binom{12}{5}=792</math> such paths. Since the path is <math>12</math> units long, they must meet after each travels <math>6</math> units, so the probability is <math>\frac{792}{(2^{6}-1)\cdot (2^{6}-1)} \approx 0.20 \Rightarrow \boxed{C}</math> |
== See Also == | == See Also == | ||
Revision as of 17:39, 11 February 2018
Problem
Objects 
and 
move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object starts at 
and each of its steps is either right or up, both equally likely. Object starts at 
and each of its steps is either to the left or down, both equally likely. Which of the following is closest to the probability that the objects meet?
Solution
If and meet, their paths connect and 
There are 
such paths. Since the path is 
units long, they must meet after each travels 
units, so the probability is 
See Also
| 2003 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 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. 
