chance of more than $m$ failures

Given that the chance of success on an attempt is $11/30$, what is the chance of more than $m$ failures? I think I should consider the complement of the statement, but I am stuck at this point.

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Do you mean "more than $m$ failures before the first success"? – Espen Nielsen Nov 20 '12 at 21:32
@ espen180 yes indeed – Badshah Nov 20 '12 at 21:33

We are concidering each trial to be independent. Then this is an example of a negative binomial distribution. See http://en.wikipedia.org/wiki/Negative_binomial_distribution

Let $p$ be the probability of success. Then the probability of getting exactly $k$ failures before the first success is given by $P_k = (1-p)^k p$, the probability of $k$ failures and one success. Now, we want the total probability for all $k>m$. This is

$$P=\sum_{k=m+1}^{\infty} (1-p)^k p$$

Do you see why this is? Can you now evaluate the probability?

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you could write P as the sum from k=0 to infinity minus the sum from k=0 to k=m. Is this correct? – Badshah Nov 20 '12 at 21:44
Yes, you can do this. However, the sum from $0$ to $\infty$ is 1 because it is a probability distribution. Nevertheless, it is a good exercise to calculate it. – Espen Nielsen Nov 20 '12 at 21:48
@ espen180 okey, thank you very much – Badshah Nov 20 '12 at 21:49

To have more than $m$ failures before the first success, you need to start with $m+1$ failures, and if the probability of success is $p$ and each failure is independnet of the others, the probability is $$(1-p)^{m+1}.$$

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isn't this the probability of precisely m+1 failures? – Badshah Nov 20 '12 at 21:49
It is the probability of exactly $m+1$ failures in the first $m+1$ attempts, followed by anything. And that is what you want. It is the same as espen180's expression. – Henry Nov 20 '12 at 21:52
@ Henry okey, now I understand. Thanks – Badshah Nov 20 '12 at 21:53