Multiple Probability?

Question

There are a box of three different types of cookies. 12 are chocolate, 8 are vanilla, and 5 are peanut butter. If Sara reaches in and grabs 3 at random, without putting them back, what is the probability of the following?

P(1 Chocolate, 1 Vanilla, 1 Peanut Butter)
P(1 Peanut Butter, 2 Vanilla)

Answer 1

P ( 1C, 1V and 1P) in that order
= 12/25 * 8/24 * 5/23 = 4/115
If the question said you needed choc, van and pea in that order, then that's your answer, but if order is not important, then there are 3! = 6 ways in which you can choose these three flavors.
So the final answer is 24 / 115

AGAIN the prob of 1P and 2V in that order
= 5/25 * 8/24 * 8/23 = 8/345
There are C(3,1) = 3 ways of choosing them
Answer = 8/115

Answer 2

P(1 Chocolate, 1 Vanilla, 1 Peanut Butter) = 12/25 x 8/24 x 5/23. These are the probabilities for each type of cookie. For example, the first one is 12/25 because there are 25 cookies total and 12 of them are chocolate. the number on the bottom goes down because after you take one cookie out, there are only 24 cookies.

P(1 Peanut Butter, 2 Vanilla) = 5/25 x 8/24 x 7/23. This one is a lot the same, but for the second vanilla, there are only 7 vanilla cookies left, so you use 7 in the numerator.

Answer 3

1. (12/25)(8/24)(5/23)
2. (5/25)(8/24)(7/23)

Answer 4

(12x8x5)X(4x8c2)/(25c3x22c3)

if she makes two draws without putting back.