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.

Question

btaylor · Answer

So far I have the probability that the different gate is less than or equal to 400 feet away from each indiv. gate.
Gate 1: 4/11
Gate 2: 5/11
Gate 3: 6/11
Gate 4: 7/11
Gate 5: 8/11
Gate 6: 8/11
Gate 7: 8/11
Gate 8: 8/11
Gate 9: 7/11
Gate 10: 6/11
Gate 11: 5/11
Gate 12: 4/11

btaylor · Answer

Since each gate has an equal chance of being chosen as the first one, do I divide each probability by 12?

btaylor · Answer

Then, I would have $$\frac{4+5+6+7+8+8+8+8+7+6+5+4}{11 \times 12}$$

btaylor · Answer

Is that right?

hoblos · Answer

well it seems logical to me.. but you want get m & n prime positive integers :s

btaylor · Answer

it simplifies to 76/132, which goes to 19/33. These are relatively prime (no common factors).

hoblos · Answer

ohh "relatively" I missed that ! you're right then 
nice job :)

btaylor · Answer

thanks!