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 nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let Di​ be the event that the original departure gate was i, and Ni​ be the event that the new gate is i. Then
===​P( distance ≤400ft)i=1∑4​P(Di​)P(N1​ through Ni+4​)+i=5∑8​P(Di​)P(Ni−4​ through Ni+4​)+i=9∑12​P(Di​)P(Ni−4​ through N12​)2⋅121​(114​+115​+116​+117​)+121​(4⋅118​)3311​+338​=3319​,​
and m+n=52​.
The problems on this page are the property of the MAA's American Mathematics Competitions