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Probability proving the series of a pmf converges to a probability.

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data

    There are two separate series I'm having trouble with, although they're related.

    The scenario: Roll a fair die until a six comes up.

    pmf = (5/6)^(x-1) * (1/6)

    So first, show the sum from 1 to infinity of p(x) =1

    Next, determine P(X=1,3,5,7,...) that it will appear on an odd roll.


    2. Relevant equations

    I'm trying to prove 1=sum (5/6)^(x-1) * (1/6)

    and sum (5/6)^(2x-1) * (1/6) = ?


    3. The attempt at a solution

    For the first, I tried induction but got lost along the way.

    The second, the back of the book states = 6/11 but getting there sends me back to computing the first series.
     
  2. jcsd
  3. Sep 24, 2009 #2

    Dick

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    Your first series is a geometric series. What's the sum of 1+a+a^2+a^3+... for a<1? The second series will be a geometric series as well with a different ratio.
     
  4. Sep 24, 2009 #3
    I'm getting sum (1/5)*(5/6)^n

    So,

    1 = sum (1/5)*(5/6)^n
    1 = (1/5) sum (5/6)^n
    5 = sum (5/6)^n
    5 = (5/6) + (5/6)^2 + (5/6)^3 + ... + (5/6)^n
    5 = (5/6)[1 + (5/6) + (5/6)^2 +...+ (5/6)^(n-1)]

    Now, I need to show this for n+1.

    5 = (5/6)[1 + (5/6) + (5/6)^2 +...+ (5/6)^(n-1)]+(5/6)*(5/6)^(n-1+1)
    5 = (5/6)[1 + (5/6) + (5/6)^2 +...+ (5/6)^(n-1)]+(5/6)*(5/6)^n
    = 5 + (5/6)(5/6)^n
    = 5 + (5/6)^(n+1)
    so true for all n.

    I'm not sure if that was correct? But I need a derivation instead of the geometric assumption.
     
  5. Sep 24, 2009 #4

    Dick

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    You don't need induction. Just sum the geometric series. It's not an assumption. It's a fact. Can you write something a little clearer next time?
     
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