5
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ How many choices of six non-yellow balls are there? How many of those choices have exactly $2$ blue balls? $\endgroup$
    – lulu
    Sep 2, 2022 at 20:46

1 Answer 1

4
$\begingroup$

As noted by @lulu -- all of these problems boil down to counting.

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.

So you have the numerator correct, but since they used "given no yellow balls were chosen" we needed to adjust the denominator.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .