# Thread: Proof of probability of being chosen

1. ## Proof of probability of being chosen

This problem is from Mathematical Statistics with Resampling and R, chapter 1. The answer in the back of the book just gives n/N which is the result they're asking you to prove (!).

The probability of any subset of size n from a population of size N is 1/(N choose n). Show that this implies that any one person in the population has a n/N probability of being chosen in a sample.

I'm not sure how to attack this. I started with n=1 and ended up with (N-1)!/N! = 1/N, and I was thinking then maybe I could use induction or something? So far no luck. Any suggestions?

2. ## Re: Proof of probability of being chosen

Let's say we're concerned with the probability that Joe will get selected. Out of the subsets - how many of them contain Joe? This is a simple counting argument. Well we need to make sure that Joe is in the subset - after that we just need to choose n-1 people from the remaining N-1 people to get the number of subsets that Joe is a part of. Can you see where it goes from here?

3. ## The Following User Says Thank You to Dason For This Useful Post:

gaeborak (08-13-2013)

4. ## Re: Proof of probability of being chosen

Ok, yeah I see it now (it's always so obvious in retrospect). Here's what I got:

5. ## Re: Proof of probability of being chosen

Looks good. One could always make an argument based on symmetry which is intuitive but I think this way is a little bit more clear.

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