Probability of N or more from uniform distribution

In summary: 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.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]
  • #1
belliott4488
662
1
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.
 
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  • #2
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
mathman said:
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.
Thank you! That's very helpful.

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 anyone 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
Yes. All that's needed is independent uniformly distributed random variables. Dimension of the space is irrelevant.
 

Related to Probability of N or more from uniform distribution

1. What is the probability of getting N or more from a uniform distribution?

The probability of getting N or more from a uniform distribution depends on the specific values of N and the range of the uniform distribution. It can be calculated by dividing the number of values in the range that are greater than or equal to N by the total number of values in the range.

2. How is the probability of N or more from a uniform distribution different from the probability of exactly N?

The probability of N or more from a uniform distribution includes all values that are greater than or equal to N, while the probability of exactly N only includes the specific value of N. This means that the probability of N or more is always equal to or greater than the probability of exactly N.

3. Can the probability of N or more from a uniform distribution be greater than 1?

No, the probability of N or more from a uniform distribution cannot be greater than 1. This is because the total probability for any event or range of values cannot exceed 1, as it represents the entire sample space.

4. How does the range of a uniform distribution affect the probability of N or more?

The range of a uniform distribution can greatly impact the probability of N or more. A wider range will result in a higher probability of N or more, while a narrower range will result in a lower probability. This is because a wider range provides more values that can be equal to or greater than N.

5. Can the probability of N or more from a uniform distribution be negative?

No, the probability of N or more from a uniform distribution cannot be negative. It is always a non-negative value, as it represents the likelihood of an event occurring. A negative probability would indicate that the event is impossible.

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