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If Y has a geometric distribution with success probability .3, what is the largest value, y0, such

that P(Y > y0) ≥ .1?

Attempt:

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.

Thanks!

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# Homework Help: Geometric Distribution problem

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