1. ## Geometric distribution problem

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

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 ?

2. Originally Posted by electricalphysics
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

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 ?
$\displaystyle \Pr(Y > y_0) \geq 0.1 \Rightarrow 1 - \Pr(Y \leq y_0) \geq 0.1 \Rightarrow \Pr(Y \leq y_0) < 0.9 \Rightarrow (0.7)^{y_0} > 0.1$ (see cdf given here Geometric distribution - Wikipedia, the free encyclopedia).

So your job is to find the largest integer solution to $\displaystyle 0.7^{y_0} > 0.1$. I suggest using trial and error as an effective approach.