Difference between revisions of "2010 AIME II Problems/Problem 4"
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There are <math>12 \cdot 11 = 132</math> possible situations (<math>12</math> choices for the initially assigned gate, and <math>11</math> choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most <math>400</math> feet apart. | There are <math>12 \cdot 11 = 132</math> possible situations (<math>12</math> choices for the initially assigned gate, and <math>11</math> choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most <math>400</math> feet apart. | ||
| − | If we number the gates <math>1</math> through <math>12</math>, then gates <math>1</math> and <math>12</math> have four other gates within <math>400</math> feet, gates <math>2</math> and <math>11</math> have five, gates <math>3</math> and <math>10</math> have six, gates <math>4</math> and <math>9</math> have have seven, and gates <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math> have eight. Therefore, the number of valid gate assignments is < | + | If we number the gates <math>1</math> through <math>12</math>, then gates <math>1</math> and <math>12</math> have four other gates within <math>400</math> feet, gates <math>2</math> and <math>11</math> have five, gates <math>3</math> and <math>10</math> have six, gates <math>4</math> and <math>9</math> have have seven, and gates <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math> have eight. Therefore, the number of valid gate assignments is <cmath>2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76</cmath>, so the probability is <math>\frac{76}{132} = \frac{19}{33}</math>. The answer is <math>19 + 33 = \boxed{052}</math>. |
== See also == | == See also == | ||
Revision as of 12:57, 26 February 2016
Problem
Dave arrives at an airport which has twelve gates arranged in a straight line with exactly 
feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks 
feet or less to the new gate be a fraction 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
There are 
possible situations (
choices for the initially assigned gate, and 
choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most feet apart.
If we number the gates 
through , then gates and have four other gates within feet, gates 
and have five, gates 
and 
have six, gates 
and 
have have seven, and gates 
, 
, 
, 
have eight. Therefore, the number of valid gate assignments is ![\[2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76\]](http://latex.artofproblemsolving.com/6/6/6/666190accdcfad0898a3b5307e998df9034b94ce.png)
, so the probability is 
. The answer is 
.
See also
| 2010 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
