# airport gates

#### veronica1999

##### Member
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
m/n , where m and n are relatively prime positive integers. Find m + n .

I got the answer 52 but I am not that happy with my solution.
(I think I solved it using brute force because the problem was simple)
Can someone show me a better way?

Total choices are 12X11.
I listed all the gates and tried each one out.
4+5 +6 + 7 +8 + 8 +8 +8 +7 + 6 +5 +4 = 76

76/121 = 19/33

Thanks.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar The horizontal and the vertical axes represent the first and the second random gates, respectively. The surrounded area represents points (x, y) such that |x - y| ≤ 4. This so because |x - y| ≤ 4 ⇔ -4 ≤ x - y ≤ 4 ⇔ x - 4 ≤ y ≤ x + 4. You need to calculate the number of thick dots. The number of unmarked points on the grid is 2(7 + ... + 1) + 12 = 7 * 8 + 12 = 68, so there are indeed 12 * 12 - 68 = 76 thick dots. The only remark is that 19 / 33 = 76 / 132, not 76 / 121.