Chance of something being fixed I'm fixing a software defect that occurs 1 in n test runs.  If I want to know that the probability of it being fixed is >= p for some 0 <= p < 1, how many times, m, do I need to run the test successfully (without the defect occurring)?
 A: I guess $m>\log(1-p)/\log(1-1/n)$ works since the probability of a faulty system running 
$m$ times without defect is $(1-1/n)^m$ and this should be smaller than $1-p$.
This seems to be a homework type question (and moreover an easy one) rather than a MO-question. 
A: If your problem was a little bit more difficult (roland-bacher already provided an easy, precise and correct solution) and your $n$ is big, you can also approximate the binomial distribution by a Poisson distribution. Repeating the test $m$ times gives then the parameter $\lambda = \frac{m}{n}$ and your goal is that $m$ is big enough that $e^{-\lambda}\ge 1-p$. So $m \ge -n\ln(1-p)$. The approximation by the Poisson distribution is pretty good, for $n = 500$ and $p = 0.99$ it yields $m > 2302.585$ instead of the correct $m > 2300.28$ given by roland-bacher's formula.
A: According to my statistics final which I took yesterday, the answer should be 
$m=\lceil 2\left(1-\frac{1}{n}\right)\text{InverseErf}^2[1-p]\rceil$ where InverseErf[x] is the Inverse Error Function.
