The probability $P'$ of getting at least $k$ successes in $n$ independent tries, given probability of a single success $s$, equals one minus the summed probabilities of getting only $0$ to $k-1$ successes:
$P'(k, n, s) = 1 - \sum\limits_{i=0}^{k-1} P(i, n, s)$
where the probability $P$ of getting exactly $k$ successes is:
$P(k, n, s) = \,_nC_k \cdot s^k \cdot (1 - s)^{n - k}$
Now suppose I want to know how many tries I need to achieve a given probability $P'$. How do I solve for $n$?
This question is a lot like the question here:
On the Total Number of Tries Required to have $n$ Successes
But in that question each trial is not independent, since it's about selecting stones from a bag without replacement. In my question, each trial is independent.