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Circular Reasoning (Posted on 2021-02-23) Difficulty: 4 of 5
A certain road has a path of a perfect circle with a single entrance/exit. A woman enters the road and walks the full circumference at a constant speed without stopping or changing direction. During her time on the road, N cars, each at its own random time during the duration of the walk, enter the circle. Each car proceeds, on the shortest path, to its own randomly selected stopping point on the circle. If cars travel 10 times as fast as the woman walks, answer the following:

1) For N=1, what is the probability that the woman “encounters” a car?

Definition: An “encounter” is when a moving car either overtakes the woman in the same direction or passes her while going in the opposite direction. If a car is stationary, there can be no encounter.

2) What is N such that there is at least a 75% chance of encountering a car?

3) For N=20, what is the expected number of encounters?

No Solution Yet Submitted by Kenny M    
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Solution solution | Comment 1 of 3
Let the woman be walking at a speed of 1 circumference per hour and the car at 10 circumferences per hour.

1) If one given car is moving in the same direction as the woman, in order to pass the woman, the woman must not have passed the car's destination before the car.  The car's destination could be anywhere from the entrance to 0.5 of the circumference.

If the car's destination is all the way at d = 1/2 circum, the woman arrives there at time t = .5, so the car has to get there by that time. As the car takes .05 time units it can have entered up until t=.45. If the destination were earlier in the woman's walk, the car could have had to enter sooner, being at the limit of immediately at the entrance, having to get there at the same time as the woman, t = 0. The relationship is linear so the average is that the car had to enter before t=.225. 

It is equally likely that the destination would be in the second half of the woman's walk, so that the car would be going in the opposite direction to the lady. If, again, the destination were at the far side from the entrance, the lady would have gotten there at t=.5, so this time the car must get there by that time or later, rather than before. The car takes .05 units of time to get there so it must have entered at t=.45 or later. If at the other extreme the destination were very near the entrance, the car would have had to enter at just about t=1. This averages to .725. In this case we need to subtract from 1, the end of the time interval, giving .275 as the length of the time interval.

As it's equally likely the destination is in the beginning half or the ending half of the woman's circuit, the probability it the mean of the two conditional probabilities, or 0.25 = 1/4.

2) (.75)^N <= .25  (prob. of no cars)

   N*log(.75) <= log(.25)
   N >= log(.25)/log(.75) = 4.8188...
   Seek N = 5; prob no cars = .237
           ; prob at least one car = .763
3) For N=20, 5 cars are expected.          

Apparently these are independent of the ratio of the car speed and the woman's speed, as the difference of one side or the other of the circle merely accounts for an equal probability change one side or the other of 1/4, as the result above is the same as for a car of infinite speed.

  Posted by Charlie on 2021-02-23 09:45:48
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