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Independent trials

  1. Apr 15, 2010 #1
    Say I have an urn containing a set of balls numbered 1 through N. I then make n selections from the urn with replacement. Thus each selection is independent.

    Let the event: [tex]A_{i}[/tex]
    [tex]i=1,2,3....,N[/tex]

    be that the ball numbered i was not chosen in the n selections.

    Then:

    [tex]P(\mathbf{A_{i}})=\left(\frac{N-1}{N}\right)^{n}[/tex]

    and [tex]P(\mathbf{A_{i}A_{j}})=\left(\frac{N-2}{N}\right)^n}[/tex] and so on....

    But aren't the events [tex]A_{i}[/tex] [tex]i=1,2,3....,N[/tex] independent as well?

    So shouldn't the probability of
    [tex]P(\mathbf{A_{i}A_{j}})=P(\mathbf{A_{i}})P(\mathbf{A_{j}})=\left(\frac{N-1}{N}\right)^{2n}=\left(\frac{2(N-1)}{N}\right)^{n}[/tex]
     
    Last edited: Apr 15, 2010
  2. jcsd
  3. Apr 16, 2010 #2
    You have just proved that the events [itex]A_i[/itex] and [itex]A_j[/itex] are not independent.

    If they were then the probability of no colours being chosen would be

    [tex](\frac{N-1}{N})^{nN}[/itex]

    but this exprssion is obviously not 0.

    (Intuitively failing to draw one colour increases the chances of drawing each of the others.)
     
  4. Apr 16, 2010 #3
    Ah ok. Thanks!
     
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