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. 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}\cdot 2^{6}} \approx 0.20 \Rightarrow \boxed{C}</math>. |
== See Also == | == See Also == |
Revision as of 20:38, 15 March 2019
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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