Question

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 $m + n.$

答案

Solution:

Number the gates $1$ through $12.$ All $12 \cdot 11 = 132$ ordered pairs of distinct (old, new) gates are equally likely, and Dave walks $400$ feet or less exactly when the gate numbers differ by at most $4.$

A gate $i$ has $\min(i + 4, 12) - \max(i - 4, 1)$ qualifying new gates: gates $1$ and $12$ have $4$ each, gates $2$ and $11$ have $5,$ gates $3$ and $10$ have $6,$ gates $4$ and $9$ have $7,$ and gates $5$ through $8$ have $8$ each. The total is $2(4 + 5 + 6 + 7) + 4 \cdot 8 = 76.$

The probability is $\frac{76}{132} = \frac{19}{33},$ so $m + n = 19 + 33 = 52.$

2010 AIME II Problem 4

Solution:

Problem 4 in Other Years