2010 AIME II Problems/Problem 4
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 .
Solutions
Solution 1
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 so the probability is . The answer is .
Solution 2
As before, derive that there are possibilities for Dave's original and replacement gates.
Now suppose that Dave has to walk feet to get to his new gate. This means that Dave's old and new gates must be gates apart. (For example, a foot walk would consist of the two gates being adjacent to each other.) There are ways to pick two gates which are gates apart, and possibilities for gate assignments, for a total of possible assignments for each .
As a result, the total number of valid gate arrangements is and so the requested probability is for a final answer of .
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 |
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