Find a geometric model, the overall corresponding area is 24 * 24 = 576, and the corresponding area for at least one of the two ships to wait when berthing is 24 2-(24-6) 2 = 252, so the probability for at least one of the two ships to wait when berthing is
252/576=7/ 16?
Set the arrival time of ship A as X and the arrival time of ship B as Y. ..
Li Lian0 < x & lt24,<y & lt24, | x-y | <; 6,
Draw a picture. According to the linear programming, the required graphic area is 252 and the total area is 576.
So the probability is 252/576=7/ 16.
The waiting time is 6 hours.
The time interval is 24 hours.
Therefore, it can be assumed that "a to x, b to y" corresponds to points {(x, y)|24≥x≥0, 24≥y≥0}.
The necessary and sufficient condition for two ships to meet is 6≥|x-y|.
If a rectangular coordinate system is established on a plane, all possible results of (x, y) are squares with a side length of 24.
The area not within 6≥|x-y| is 18? =324?
So the probability that at least one of these two ships will have to wait when berthing is 324/24? =9/ 16.