Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 mistake.now I corrected it. –  user58267 Jan 15 '13 at 6:33
add comment

3 Answers

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.

share|improve this answer
add comment

$p=0$ 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.

share|improve this answer
    
@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
add comment
  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
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.