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Mathematical model of swimmer's probability of returning to shore. Please write in more detail.
Assuming that the river bank is in the east-west direction, the west is in the positive direction, and the starting point is O. After swimming for 200 meters, it can be assumed that the directed angle between OA and the positive direction of the river bank is A, then A obeys the uniform distribution of (0, 180) U (0, 180), OA=200, and the vertical distance from A to the river bank is: 200.

Let the included angle from point A to the positive direction be: b, then b also obeys the uniform distribution of (0, 180) u (0, 180), the new distance is: 200sin(a)/sin(b), and the probability of the problem is:

200 sin(a)/sin(b)& lt; The probability of 200 is recorded as p (200 sin (a)/sin (b).

p(200 sin(a)/sin(b)& lt; 200)= P(sin(a)/sin(b)& lt; 1)= P(sin(a)& lt; Sin(b)). Since A and B obey the same distribution, the distribution of sin(a) and SIN (b) is exactly the same, so the above probability must be: 0.5.