Difference between revisions of "2003 AMC 12A Problems/Problem 22"
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== 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, 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}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}</math> |
== See Also == | == See Also == |
Revision as of 14:01, 28 November 2013
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 |
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