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Summation of infinite series

  1. Feb 3, 2008 #1
    I'm having trouble picking apart this summation:

    [tex]\sum[/tex][tex]^{inf}_{n=1}[/tex] P(E)*P(1-p)[tex]^{n-1}[/tex]; where p = P(E) + P(F)

    I know I need to use the identity of a geometrical series when |r| < 1 : 1/(1-r)

    I'm getting [tex]P(E)/(1-(P(E)+P(F))[/tex]

    But I need to be getting P(E)/((P(E)+P(F));

    The entire problem is

    Let E & F be mutually exclusive events in the sample space of an experiment. Suppose that the exp is repeated until either event E or F occurs. What does the sample space of hte new super experiment look like? Show that the probability of event E before event F is P(E)/(P(E)+P(F)).
  2. jcsd
  3. Feb 4, 2008 #2


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    That's a bit confusing. If p= P(E)- P(F), then what is just P by itself? Asuming your terms are just A(1-p)n-1, then I would be inclined to first change the summation index: let i= n-1 so this becomes
    [tex\\sum_{i=0}^\infty A(1-p)^i[/tex]
    That sum is, of course,
    [tex]\frac{A}{1- (1-p)}= \frac{A}{p}= \frac{P(E)*P}{P(E)+ P(F)}[/tex]
    because 1-(1-p)= p. I can't say more because I am still not sure what "P" is.

  4. Feb 4, 2008 #3
    Its a probability question, so P(1-p) is just the probability that p has not occurred yet, where p = P(E U F). Since E,F are mutually exclusive, can say p = P(E) + P(F).
  5. Feb 4, 2008 #4
    But you pretty much answered my question...naturally it was a mistake in arithmetic. Thanks for helping me with that...sometimes one gets so wrapped up in the global outcome of the problem and loses track of the little details.
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