# Balls with and without replacement

I am trying to teach myself certain topics in this field. I came across this question and wanted to know if it was correct.

A bag contains $2$ red, $3$ green, and $4$ black balls. If five balls are drawn in succession, with replacement after each draw, what is the probability of getting $2$ red, $2$ green, and $1$ black ball?

My answer:

$\frac{5!}{2!*2!*1!}(2/9)^2(3/9)^2(4/9)$

=$.0731$

What if it was drawing without replacement?

My answer:

$\binom{2}{2}\binom{3}{2}\binom{4}{1}/ \binom{9}{5}$

$=.0952$

Can someone please tell me if this is correct? Is there other ways of thinking about this?

• I have not checked the arithmetic. There are missing parentheses in the first answer. The (implicit) reasoning that was used is correct. – André Nicolas May 12 '14 at 1:12
• @AndréNicolas Thank you! – earl May 12 '14 at 1:15
• You are welcome. – André Nicolas May 12 '14 at 1:24

## 1 Answer

In the no replacement, presumably the $2! * 2! *1!$ should have parentheses around them to be in the denominator. The $4/9$ should not be squared. But I agree with your final numeric answer. The without replacement is fine.

• Thank you!. Some of it were typo errors. – earl May 12 '14 at 1:18