Q: Permutations and Combinations In a group of 30 ball bearings, 5 are defective. If 10 ball bearings are chosen at random, a) what is the probability that none of them are defective? b) what is the probability that two or more defective ball bearings are chosen?
I know this has to do with a permutation problem and for part a) I got $(\dfrac{25}{30})^{10}$ not sure if i calculated correctly. I'm not sure for part b) but i got  $\dfrac{1}{6} * (\dfrac{5}{6})^9$

A: This is a hypergeometric distribution, not a binomial distribution. So if you want no defective balls, see below. You choose $10$ working balls from the $25$ working balls, and no balls from the defective title.
$$ \dfrac{ \binom{25}{10} * \binom{5}{0} }{ \binom{30}{10} }$$
For (b), just take $1 - P(X \leq 1)$. So how many ways do you get $0$ defective balls? How many ways do you get one defective ball?
If you want a single defective ball, you have probability:
$$ \dfrac{ \binom{25}{9} * \binom{5}{1} }{\binom{30}{10}}$$
Notice getting exactly one defective ball is disjoint from getting no defective balls; so by rule of sum, you add the two probabilities together. This gives you the complement of at least $2$ defective balls.
