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Math Help - Probability

  1. #1
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    Probability

    p(k) denotes the probability of exactly k successes in n independent trials of a random experiment in which the probability of a success is p. q=1-p
    How do you show that :
    p(k+1) = (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q
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  2. #2
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    Quote Originally Posted by Pink Lady View Post
    p(k+1) = (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q
    Can you tell me what the above is.
    I really have no idea what is could possibly mean.
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  3. #3
    MHF Contributor matheagle's Avatar
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    p(k) = {n\choose k}p^kq^{n-k}

    so I guess you need to play with

    p(k+1) = {n\choose k+1}p^{k+1}q^{n-k-1}

    But I'm not too sure what this.... (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q.... is
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  4. #4
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    It was meant to be written as follows:
    p(k+1) / p(k) = (n-k)p / (k+1)q , when k=0,1,2,3,...,n-1 and q=1-p

    Sorry about the confusion
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  5. #5
    MHF Contributor matheagle's Avatar
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    Well, that follows directly from what I wrote.
    Just divide the two and use x!/(x-1)!=x twice.
    AND it's k=0,1,..,n not n-1.
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