To calculate the probability of 'at least' in Binomial distributions, you have to find the sum which I'm trying to find a simplified form of $$ \sum_{i=k}^n \binom{n}{k} x^k (1-x)^{n-k}, $$ where $x \in [0,1]$ and $n,k \in \mathbb{N}$.
Maybe this has been done. I have had no luck and I don't know the terminology to request for it directly.
Presumably you mean $x^i(1-x)^{n-i}$ inside the sum. In that case, I think there is no nicer formula, except in special cases, such as $k=0$ or $k=n$ or $x=1/2$.
Maple evaluates this in terms of a hypergeometric; but it is really just the definition of the series... $$ \sum _{i=k}^{n}{n\choose i}{x}^{i} \left( 1-x \right) ^{n-i} ={{n\choose k}\frac {{x}^{k}}{ \left( 1-x \right) ^{-n+k}} \;{\mbox{$_2$F$_1$}\left(1,-n+k;\,k+1;\,{\frac {x}{-1+x}}\right)}} $$