Suppose we draw a dice 300 times.
What is the probability to get '1' or '2' in more then 90 draws?
So $X\sim Bin(300, \frac{1}{3})$ ($X$ binomially [if that even a word] distributed with parameters $300$ and $\frac{1}{3}$), and we want to find $\mathbb{P}(X>90)=1-\mathbb{P}(X\leq 90)$.
$$1-\mathbb{P}(X\leq 90)=1-\sum_{k=0}^{90}\binom{300}{k}p^kq^{300-k}=$$ $$1-q^{210}\sum_{k=0}^{90}\binom{300}{k}p^kq^{90-k}$$ (where $p=\frac{1}{3}$ and $q=\frac{2}{3}$)
But now I'm not really sure how to proceed.
Is there a way to take the $\binom{300}{k}$ to the form of $\binom{90}{k}$ somehow, so I will be able to use the Binomial theorem?
