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Math Help - [SOLVED] questions are about sampling.

  1. #1
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    Exclamation [SOLVED] questions are about sampling.

    Can someone help me with the following:

    see the attachment.

    I am really stuck and do not know were to start

    Thanks

    Maths_Brain
    Attached Thumbnails Attached Thumbnails [SOLVED] questions are about sampling.-gash.png  
    Last edited by CaptainBlack; October 17th 2008 at 09:58 AM.
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  2. #2
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    Quote Originally Posted by maths_brain View Post
    Can someone help me with the following:


    see the attachment.

    I am really stuck and do not know were to start

    Thanks


    Maths_Brain
    You might find that many members are reluctant to open attached files (I'm one of those members). So you might get a better response if you take the time and type out your questions.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by mr fantastic View Post
    You might find that many members are reluctant to open attached files (I'm one of those members). So you might get a better response if you take the time and type out your questions.
    Having been stupid enough to open said attachment what do I find? It's not a word document its a wrapper for picture!! Grrrr...

    CB

    (I'm not vulnerable to macro virus's but I don't like opening word and excel attachments because it takes a long time to boot up OOO on my slow computers (I won't open an office document on the my work computers as a matter of principle)
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  4. #4
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    Quote Originally Posted by maths_brain View Post
    Can someone help me with the following:

    see the attachment.

    I am really stuck and do not know were to start

    Thanks

    Maths_Brain
    The third part is just algebra, write out the right hand side in terms of factorials and simplify.

    <br />
{n-1 \choose r}+{n-1 \choose r-1}=\frac{(n-1)!}{r!(n-1-r)!} + \frac{(n-1)!}{(r-1)!(n-r)!}<br />

    Alternativly one may observe that it is the sum of the number of ways of choosing a subset of r elements from the given set which does not include 1, and the number of ways that does include 1.

    CB
    Last edited by CaptainBlack; October 17th 2008 at 10:22 AM.
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  5. #5
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    Quote Originally Posted by maths_brain View Post
    Can someone help me with the following:

    see the attachment.

    I am really stuck and do not know were to start

    Thanks

    Maths_Brain
    When choosing a subset consisting of r elements from a set of n elements the first element may be choosen in n ways the second in (n-1) ways and so on to the r-th which may be choosen in (n-r+1) ways. So the number of ways that the set may be choosen is n(n-1)...(n-r+1), but in this number every permutation of each distinct choice appears, so to get the number of choices independent of order we divide by the number of permutations of r distinct objects. Hence the number of ways of choosing the subset is:

    N(r,n)=\frac{n(n-1)...(n-r+1)}{r!}=\frac{n!}{r!(n-r)!}

    CB
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  6. #6
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    Quote Originally Posted by maths_brain View Post
    Can someone help me with the following:

    see the attachment.

    I am really stuck and do not know were to start

    Thanks

    Maths_Brain
    When choosing a subset consisting of r elements from a set of n elements the first element may be choosen in n ways the second in (n-1) ways and so on to the r-th which may be choosen in (n-r+1) ways. So the number of ways that the set may be choosen is n(n-1)...(n-r+1), but in this number every permutation of each distinct choice appears, so to get the number of choices independent of order we divide by the number of permutations of r distinct objects. Hence the number of ways of choosing the subset is:

    N(r,n)=\frac{n(n-1)...(n-r+1)}{r!}=\frac{n!}{r!(n-r)!}

    Which is what we expect.

    Now the number of ways of choosing a subset that does not contain 1 is N(r,n-1), and so the probability required is:

    p=\frac{N(r,n-1)}{N(r,n)}

    Similarly if 1 is in the set the number of ways of choosing it is N((r-1),(n-1))

    CB
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  7. #7
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    Thanks for the help Captin Black.

    Just wondering is the probaility of it containing 1 in the set equal to N((r-1),(n-1))

    and the number of ways of choosing a subset that does not contain is N(r,n-1)

    Thanks

    Maths Brain
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  8. #8
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    Quote Originally Posted by maths_brain View Post
    Thanks for the help Captin Black.

    Just wondering is the probaility of it containing 1 in the set equal to N((r-1),(n-1))

    and the number of ways of choosing a subset that does not contain is N(r,n-1)

    Thanks

    Maths Brain
    If you read what I said earlier you will realise that N((r-1),(n-1)) is the number of ways of choosing the set with a 1 in it, to get the probability you have to divide by the total number of sets (N(r,n))

    (same for the other one)

    CB
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