Probability of picking at least x blue balls out of red and blue balls Given $n$ balls in total.
Each ball is either red or blue.
There are $ r $ red balls and $ b $ blue balls.
At random, and without replacement, we pick $ t $ balls out of the $ n $ balls.
How to calculate the probability for the event that at least $ x $ of the picked balls are blue?
For an example consider: $n=5000$, $r=1500$, $b=3500$, $t=500$, $x=350$

The probability of first getting $x$ blue balls, followed by $t-x$ red balls, equals:
$$\frac{b}{n} \cdot \frac{b-1}{n-1} \cdot \ldots \cdot \frac{b-x+1}{n-x+1} \cdot \frac{r}{n-x} \cdot \frac{r-1}{n-x-1} \cdot \ldots \cdot \frac{r-t+x+1}{n-t+1} = \frac{b!}{(b-x)!} \frac{r!}{(r-t+x)!} \frac{(n-t)!}{n!}$$
There are ${{t}\choose{x}}$ ways to select the turns in which we pick a blue ball, so the probability of selecting exactly $x$ blue balls equals:
$${{t}\choose{x}} \frac{b!}{(b-x)!} \frac{r!}{(r-t+x)!} \frac{(n-t)!}{n!}$$
The probability of selecting at least $x$ balls then becomes:
$$\sum_{x=350}^{500}{{t}\choose{x}} \frac{b!}{(b-x)!} \frac{r!}{(r-t+x)!} \frac{(n-t)!}{n!}$$
