Let there be $R$ red and $B$ blue balls, with each ball distinct from the other (even of the same colour). $M$ balls ($(1)$ assume $M<R,B$) are to be chosen. What is the probability that the number of red balls chosen, $n(R)$, is $\ge x$ ($x<M$)?
Uncivilised solution ($N(...)$ is the number of ways of doing $...$):
$$\large P(n(R) \ge x)=\frac{N(n(R)\ge x)}{N(n(R)< x)+N(n(R)\ge x)}$$
$$N(n(R)\ge x)=N(n(R)= x)+N(n(R)= x+1)+...+N(n(R)= M)$$ $$=\frac{R!}{x!(R-x)!}\cdot\frac{B!}{(M-x)!(B-M+x)!}+...$$ $$=\sum_{i=x}^M\frac{R!}{i!(R-i)!}\cdot\frac{B!}{(M-i)!(B-M+i)!}$$ $$=\sum_{i=x}^M\binom{R}{i}\binom{B}{M-i}$$ Similarly, $$N(n(R)< x)=\sum_{i=0}^{x-1}\binom{R}{i}\binom{B}{M-i}$$
But like all uncivilised solutions, I feel it's necessarily long and overcomplicated. Is there a better way to solve this problem? Does this way work if restriction $(1)$ doesn't apply?
It feels a little like the territory of the principle of inclusion and exclusion, but I'm not sure how to apply it here.
It is uncivilized to use caps. So assume that there are $r$ red and $b$ blue, and we are choosing $m$ balls.
There are $\binom{r+b}{m}$ equally likely ways to choose $m$ balls from the $r+b$.
The number of ways to choose $i$ red and therefore $m-i$ black is $\binom{r}{i}\binom{b}{m-i}$. So our probability is $$\frac{\sum_{i=x}^m \binom{r}{i}\binom{b}{m-i}}{\binom{r+b}{m}}.$$ There is no nice closed form for the above sum.