Probability: 30 balls in a bucket, homework i need some help with some homework, first time i am doing probability and statistics, id like to know if my 2 answers below are correct, and how i can solve the remaining 2.
There are 30 balls in a bucket:
10 green
12 red
8 blue
3 Balls are pulled up from the bucket.
a) What is the probability of all 3 being blue.
$$
\frac{8}{30} * \frac{7}{29} * \frac{6}{28} = \frac{2}{145}
$$
c) What is the probability of all 3 balls being different colors.
$$
\frac{10}{30} * \frac{8}{29} * \frac{12}{28} = \frac{8}{203}
$$
d) What is the probability of atleast 2 balls being green
e) What is the probability of 1 blue ball max.
 A: Your answers for (a) and (b) are correct. It might help to think about these combinatorially though. For (a), you choose three blue balls. That's $\binom{8}{3}$ ways of choosing your blue balls. There are $\binom{30}{3}$ possible ways to choose 3 balls. So your probability is 
$$\frac{ \binom{8}{3} }{ \binom{30}{3} }$$
It's the same answer, but a bit more condensed and self-explanatory, I think.
For (d), we get two balls being green in the following way: $\binom{10}{2}$. We then select one ball from the red + blue pile, which is $\binom{20}{1}$. We then divide out by the number of ways to get three balls, which is $\binom{30}{3}$. 
We then evaluate the number of ways to choose exactly three green balls, which is $\binom{10}{3}$. We again divide out by $\binom{30}{3}$. The number of ways to get exactly two and three balls are independent, so we add these quantities together.
For (e), we have the same approach as (d). How many ways can we get no blue balls out of three? That's $\binom{22}{3}$ right? And to get one blue ball exactly, we have $\binom{8}{1} * \binom{22}{2}$. These conditions are independent, so we add them together and divide out by the number of ways to choose three balls.
