Probability of picking buttons from a bag A bag contains $30$ buttons that are colored either blue, red or yellow. There are the same number of each color ($10$ each). A total $4$ buttons are drawn from the bag. Compute the followings:

*

*Find $n(\Omega)$.

*The probability that at least $3$ of them are red?

*The probability that there is at least one of each color?

This seems like a basic problem but my professor and I cannot agree on an answer.
I think the probabilities are $2/21$ and $100/203$ for parts $2$ and $3$ respectively. I used combinations to calculate the probabilities.
My professor said $n(A)/n(\Omega)$ is $3/15$ for both so that is the answer for both $2$ and $3$.
 A: 
A bag contains $30$ buttons that are colored either blue, red or yellow. There is the same number of each color ($10$ each). A total of $4$ buttons are drawn from the bag. Compute the following:
$1$. Find $n(\Omega)$.


$2$. The probability that at least $3$ of them are red.

$$\frac{{10\choose3}{20\choose1}+{10\choose4}}{{30\choose4}}=\frac{2610}{27405}=\frac{2\cdot3^2\cdot5\cdot29}{3^3\cdot5\cdot7\cdot29}=\frac 2{21}$$

$3$. The probability that there is at least one of each color.

$$\frac 12\times\frac{{10\choose1}{10\choose1}{10\choose1}{27\choose1}}{{30\choose4}}=\frac 12\times\frac{10\cdot10\cdot10\cdot27}{27405}=\frac 12\times\frac{2^3\cdot3^3\cdot5^3}{3^3\cdot5\cdot7\cdot29}=\frac {100}{203}$$
where the factor of $\frac12$ was to cancel double-counting, in that every combination like $(\underbrace{R_i}_{{10\choose1}}\underbrace{B_i}_{{10\choose1}}\underbrace{Y_i}_{{10\choose1}}\underbrace{R_j}_{{27\choose1}})$ is also counted in $(R_jB_iY_iR_i)$.
You are correct.
A: Part B can also be done as
$P(B) = \frac{3C1.(10C2.10C1.10C1)}{30C4} = \frac{100}{203}$.
A: I think we have the same professor, and I just received my assignment back and I basically got all of it wrong because of this question, he said that it was 15 not 27405.
