# Finding $i$ such that $\sum_{j=0}^i\binom{n}{j}\left(1-\frac1k\right)^j\left(\frac1k\right)^{n-j}\approx\frac1k$

Let $n,k$ be positive integers. From the binomial theorem, we know that

$$\left(\frac{1}{k}\right)^n+n\left(1-\frac1k\right)\left(\frac1k\right)^{n-1}+\binom{n}{2}\left(1-\frac1k\right)^2\left(\frac1k\right)^{n-2}+\cdots+\left(1-\frac1k\right)^n=1$$

For what $i$ (in terms of $n,k$) is it true that the sum of the first $i$ terms is approximately $\frac1k$? In other words, for what $i$ is it that

$$\sum_{j=0}^i\binom{n}{j}\left(1-\frac1k\right)^j\left(\frac1k\right)^{n-j}\approx\frac1k?$$

I'm not sure how to start with approximating this binomial sum.

If $n$ is large enough, such a distribution can be approximated by a normal distribution and the answer be formulated in terms of $\Phi^{-1}$, i.e. the inverse function of the CDF of a normal distribution.