HMMT 二月 2006 · 代数 · 第 3 题

英文解析

3. The train schedule in Hummut is hopelessly unreliable. Train A will enter Intersection
   X from the west at a random time between 9:00 am and 2:30 pm; each moment in thatinterval is equally likely. Train B will enter the same intersection from the north at a random time between 9:30 am and 12:30 pm, independent of Train A; again, eachmoment in the interval is equally likely. If each train takes 45 minutes to clear theintersection, what is the probability of a collision today?
   Answer:13
   Solution: Suppose we fix the time at which Train B arrives at Intersection X; then 48
   call the interval during which Train A could arrive (given its schedule) and collide with
   Train B the “disaster window.”
   We consider two cases:
   (i) Train B enters Intersection X between 9:30 and 9:45. If Train B arrives at 9:30,
   the disaster window is from 9:00 to 10:15, an interval of 1 hours. If Train B1
   arrives at 9:45, the disaster window is 1 hours long. Thus, the disaster window 14
   1 1 11 12
   has an average length of (1 + 1 ) ÷ 2 = . From 9:00 to 2:30 is 5 hours. The
   4 2 8 2
   11 1 1
   probability of a collision is thus ÷ 5 = .
   8 2 4
   (ii) Train B enters Intersection X between 9:45 and 12:30. Here the disaster window 1
   1 1 1 3
   is always 1 hours long, so the probability of a collision is 1 ÷ 5 = .
   2 2 2 11
   1 1
   From 9:30 to 12:30 is 3 hours. Now case (i) occurs with probability ÷ 3 = , and
   4 12
   case (ii) occurs with probability . The overall probability of a collision is therefore 11
   1 1 11 3 1 1 1312
   · + · = + = .
   12 4 12 11 48 4 48