Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Geometric Distribution, I am getting the probability for doing $x$ number of trials and get my first success with each trial of probability $p$.

So suppose I want to find what's the probability of doing 30 trials and get my first success on the 30th trial, I do this: $$ P(X=30)=(1-p)^{30-1}p$$

Now, then if I want to find the probability for not getting a single success at all even after doing $30$ trials on this same distribution, what should I do? The parameters of the Geometric Distribution doesn't seem to let me find this.

I thought of using like $1$ minus the CDF of 30 trials of the geometric distribution but I am not sure if it would be accurate.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

How did you arrive at $P(X=30)=(1-p)^{30-1}p$? Usually this is derived by arguing that to have the first success in the $30$-th trial you need to have $29$ trials without success and then one trial with success, which makes $(1-p)^{29}p$. But that argument already contains the answer to your question, namely, since the probability of not getting a success in one trial is $1-p$, the probability of not getting a success in $29$ trials is $(1-p)^{29}$, and analogously the probability of not getting a success in $30$ trials is $(1-p)^{30}$. So you took the second step before the first.

share|improve this answer
    
oh ya! Thanks! Should have thought of this. I didn't arrive to the equation but merely just took and used the formula. Maybe this is why I didn't see the intuition behind the formula. I did a calculation, and it turns out that if I take $1-P(X \leq 30)$, where $P(X)$ is the geometric distribution PMF, the value is the same as $(1-p)^{30}$. Which aligns with the intuition you explained. Thanks! –  xenon Feb 22 '12 at 17:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.