Finding y0 where P(Y > y0) ≥ 0.1 for geometric distribution

[FaradayLaws]

Question:
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!

 

[mathman]

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.

 

[FaradayLaws]

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 ?

 

[mathman]

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).