Probability of N or more from uniform distribution

1. Feb 3, 2009

belliott4488

If I have N objects uniformly placed at random in a 1-d box of length b, how do I calculate the probability of finding one or more objects in a given length?

Here's what I mean:

I assume a uniform probability density of 1/b, that is, P = 1/b for 0<x<b and P = 0 everywhere else. I now place N objects into this region, according to that distribution. Thus I have a distribution of objects given by N/b for 0<x<b and zero everywhere else.

This seems to make sense, i.e. my total probability is 1, then total expected number of object in the box is N, and I can calculate the expected number in any segment of length a within the box as a*N/b.

What I am having trouble with is expressing the probability that I will find at least one object in a segment of length a. If a=b, then I'd better get 1 as my answer, but I haven't come up with the correct expression.

I guess I could also ask for the probability of finding n>=2, n>=3, ... n=N, as well, each of which should approach 1 as the length approaches b.

This feels like a pretty elementary question - is it?

Thanks for any suggestions.

Last edited: Feb 3, 2009
2. Feb 3, 2009

mathman

The usual way to get an answer to the probability of at least 1 item is to find the probability of 0 items and subtract from 1. In your case the probability that a given object will not appear in an interval of length a is (b-a)/b. Therefore the probability that all n will not appear in the interval is [(b-a)/b]n. So the probability that at least 1 will be in the interval is 1-[(b-a)/b]n.

3. Feb 3, 2009

belliott4488

Now maybe I can apply that to my real problem, which is to calculate the probability of collision between an object of a given cross-sectional area and one or more particles that are distributed uniformly in a given volume, through which the object is passing. I think this should just be a matter of extending the 1-d lengths to 3-d volumes, yes?

If the object sweeps through a volume v, then the probability of any one particle not being in that volume is (V-v)/V (where V = total volume containing particles). The probability that none of the N particles is in the volume v is [(V-v)/V]N, so the probability that at least one particle is in volume v is 1 - [(V-v)/V]N.

Did I get that right?

4. Feb 4, 2009

mathman

Yes. All thats needed is independent uniformly distributed random variables. Dimension of the space is irrelevant.