# Finding the probability of getting no successes in a Geometric Distribution

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.

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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.
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