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Inside a square of side $2$ units , five points are marked at random. What is the probability that there are at least two points such that the distance between them is at most $\sqrt2$ units?

Totally stuck on it. How can I solve this.

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Exactly $\sqrt{2}$? If so, it will be 0, since the probability that any 2 points are a fixed distance apart is 0. – Calvin Lin Jan 15 '13 at 5:16
sorry for my I corrected it. – user58267 Jan 15 '13 at 6:33

You probably want to ask what is the probability that at least one pair of points is less than $\sqrt 2$ apart. If so, you could think about the pigeonhole principle.

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$p=0$ because the furthest separation between 5 points in a square is with one at each corner and one in the middle, but by Pythagoras theorem the distance between the corner and the middle is $\sqrt2$. Not a formal proof I grant you but the logic is infallible.

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@fabee that's no problem, the center of a square of side length $l$ is ${l^{2}}\over2$ – Dale M Jan 15 '13 at 9:34
  1. Divide the square into 4 squares of side length 1.In at least one square there are two or more points. All points in the same square have distance less than 2^0.5. So the answer is 1
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