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Thread: Proof of probability of being chosen

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    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?

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    Re: Proof of probability of being chosen

    Let's say we're concerned with the probability that Joe will get selected. Out of the \binom{N}{n} 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?
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    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:

    \frac{ {1 \choose 1}{N-1 \choose{n-1}}} {{N \choose n}}=
\frac{1 \times \frac{(N-1)!}{[(N-1)-(n-1)]! (n-1)!}}{\frac{N!}{(N-n)! n!}}=
\frac{(N-1)! (N-n)! n!}{N! (N-n)! (n-1)!}=
\frac{n}{N}

    Thanks for your help!

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    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.
    I don't have emotions and sometimes that makes me very sad.

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