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
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 ?