1. The first one is given by a hyper geometric distribution.

2. I think you are saying just a 13 card deck of cards. So after drawing the first card you have a chance of not drawing the same card. On the third draw you have a chance of not drawing either of the previous 2 cards. The product of the two probabilities is your answer, that is:

3. P(first ball being white) P(2nd ball being white) (tree diagram)

4. P(defective) = . So the probability its not defective is the complement of this: If the chosen product is defective, what's the probability that it was processed by machine two?(conditional probability) P(From machine 2| Defective)

5. P(choosing all new balls for the 2nd match) = P(B)

Let P(choosing all new balls for the 1st match) = P(A). We want the P(A|B) = P(AB)/P(B). But because these 2 events are independent P(A|B) = P(A) =

If someone wants to check if im right that would be nice