2010 AIME II Problems/Problem 4

Problem

Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $100$ 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 $400$ feet or less to the new gate be a fraction $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

There are $12 \times 11 = 132$ potential gate assignments. We need to count the valid ones.

Number the gates $1$ through $12$ . Gates $1$ and $12$ have four gates within $400$ feet. Gates $2$ and $11$ have five. Gates $3$ and $10$ have six. Gates $4$ and $9$ have have seven. Gates $5$ through $8$ have eight.

Therefore, the number of valid gate assignments is $2 \times (4+5+6+7+2\times8) = 76$ , and the probability is $\frac{76}{132} = \frac{19}{33}$ . Hence, the answer is given by $19 + 33 = \boxed{052}$ .

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2010 AIME II Problems/Problem 4

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