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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]

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



    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
    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


    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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