Consider an interval of length $A$ with $N$ points distributed on it according to homogenous

distribution. Each point is encircled with a interval of width $X$ (where $X \ll A$).

Let us randomly pick a new point on the interval.

What is the probability of picking a point inside at least one of the little intervals?


Take a square $A\times A$ and on it a diagonal stripe of vertical height $\pm X/2$.
If $X<<A$, the area of the stripe will be $AX$ and its ratio to the whole area $X/A$.
That's the probability that two points, uniformly distributed over $A$, be separated by a distance less than $X/2$, and for small $X$ it is indipendent from the position of the first point.

Taking the $N$ couples $(P_{N+1},\, P_1),\, \cdots,\, (P_{N+1},\, P_N)$, the probability that all of them be separated by a distance larger than $X/2$ will be $(1-X/A)^N \approx 1-NX/A$.
So the probability you are looking for will be ...

  • $\begingroup$ Thank you very much for your answer! I don't understand the diagonal stripe explanation, but it seems I understand now how one can do this. One point is chosen from range $A$, then you can chose the other from range of width $X$, in order to have low enough distance. Accesible area to get desired effect is $XA$ and whole area of all possible choices is $A^2$. Probability $X/A$ follows. Is that correct way of solving it? Second paragraph of your answer I can follow. $\endgroup$ – user535260 Feb 26 '18 at 20:20
  • 1
    $\begingroup$ @PrzemysławP: yes, it is the $(x,y)$ representation of two uniform variables over $[0,A]$. The stripe is representing the points $x-X/2 < y <x + X/2$. $\endgroup$ – G Cab Feb 26 '18 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy