# drawing balls from urn without out repetition and with repetition

So the question is that you have an urn that contains 7 red,white, and blue balls and you are going to draw a sample of 5 without replacement. What is the probability that you will draw 4 balls of one colour and one ball of another?

Now I just want to make sure that without replacement that it would be $$\frac{{3\choose1}\times{7\choose5}\times{2\choose1}\times{7\choose1}}{{21\choose5}}$$

Also I am wondering that if the above answer is correct, how would you do this if replacement was allowed? Would you just use stars and bars technique?

Assuming that you mean $$\binom{7}{4}$$ instead of $$\binom{7}{5}$$, yes, this is correct.
In the case where balls are replaced, observe that you're really dealing with a $$\frac{1}{3}$$ chance of getting any color of ball, independently, in any round. So it suffices to count the number of good permutations: $$3 \times 2 \times 5$$ (three ways to choose the first color, two ways to choose the second, and five ways to choose where the odd ball out is picked). Multiply this by the probability any particular permutation occurs, which is $$\left(\frac{1}{3}\right)^5$$, to get $$\frac{10}{81}$$.
• This is for different orders. For example, I’m counting $RRRRW$ and $WRRRR$ separately, so that the probabilities work out nicely. – platty Nov 29 '18 at 15:11