Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 3 down vote accepted

We are concidering each trial to be independent. Then this is an example of a negative binomial distribution. See

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?

share|cite|improve this answer
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 $$(1-p)^{m+1}.$$

share|cite|improve this answer
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

Your Answer


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.