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Geometric Distribution problem

  1. May 15, 2010 #1
    If Y has a geometric distribution with success probability .3, what is the largest value, y0, such
    that P(Y > y0) ≥ .1?

    So i represented the probability of the random variable as a summation

    Sum from y0= y0+1 to infinity q^(yo+1)-1 p ≥ .1
    using a change of variables i let l = y0+1

    p Sum from y0=l to inf (q)^l-1 ≥ .1

    from here i'm stuck.. i was thinking of applying the partial sum for the geometric series but i'm not sure how to proceed from here.

  2. jcsd
  3. May 15, 2010 #2


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    Science Advisor

    It is easier to work from the other end. Find max y0 such that P(Y < y0) < .9. You will then have a finite sum (geometric series) to work with.
  4. May 15, 2010 #3
    oh okay;

    once working with the other end =>

    summation from y0 =0 to y0-1 of q^y0-1 p < 0.9

    with the change of variables l= y0-1

    summation from l=0 to l of q^l p < 0.9

    now finding the partial sum of the geometric series

    p/(1-q) < 0.9
    0.3/ 0.3 < 0.9

    i'm stuck here ? how do i get the value for y0 ?
  5. May 16, 2010 #4


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    Your partial sum doesn't look right. It should be a function of y0. It should be something like 1-.3^(y0) < .9. (I am not sure whether it should be y0 or y0+1).
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