# Given that no yellow balls were chosen, what is the probability that there are exactly $2$ blue balls chosen?

I'm struggling with this problem. Can anyone tell me what I'm doing wrong?

Six balls are to be randomly chosen from $$9$$ blue, $$9$$ yellow and $$14$$ green balls. Given that no yellow balls were chosen, what is the probability that there are exactly $$2$$ blue balls chosen?

I was thinking the answer is:

$$\frac{\binom{9}{2} \binom{14}{4}}{\binom{32}{6}}$$

but this isn't right and I really don't know how else to approach it.

• How many choices of six non-yellow balls are there? How many of those choices have exactly $2$ blue balls?
– lulu
Sep 2, 2022 at 20:46

You correctly identified that there are $$32 \choose 6$$ ways to chose a set of 6 balls without replacement. However, we are told no yellow balls were chosen, so we can restrict our denominator to just the other two colors.
$$N = {23 \choose 6}$$
Of these, there are only $${9 \choose 2}{14 \choose 4}$$ ways to select exactly 2 blue balls.