Formalization/Verification of a beginner combinatorics problem I have the following task:
To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?


*

*No defective components

*Exactly one defective component

*Exactly two defective components



Progress so far:
The general approach I use is to consider the number of non-defective components over the total number of components for repeated decrements of each set.


*

*$\frac{10}{12} \cdot \frac{9}{11} \cdot \frac{8}{10} \cdot \frac{7}{9} \approx 0.4242 $
For the next question, I sum the probabilities of each case, depending on if the defective component is chosen last, second last, etc. 


*$\frac{10}{12} \cdot \frac{9}{11} \cdot \frac{8}{10} \cdot \frac{2}{9} \ + \frac{10}{12} \cdot \frac{9}{11} \cdot \frac{2}{10} \cdot \frac{8}{9} \ + \frac{10}{12} \cdot \frac{2}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} \ + \frac{2}{12} \cdot \frac{10}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} $
I could employ a similar intuitive approach for 3. , however I feel like this is messy and not particularly rigorous. How do I answer these questions formally?

The probability on $k\in\{0,1,2\}$ defectives among the $4$ chosen is:$$\frac{\binom4k\binom8{2-k}}{\binom{12}2}$$
Hypergeometric distribution is used. 
Actually this approach turns things around: there are $4$ chosen items and $8$ non-chosen items. Selecting $2$ items (the defective ones) what is the probability that $k$ of them will come from the chosen ones?
