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?

What you have so far is good. To simplify case 2, you can notice that the probabilities are all the same: for each one you get $12\times11\times10\times 9$ on the bottom and $10\times9\times8\times 2$ on the top in some order. So the total probability is just the probability for a specific "pattern" (e.g. defective, good, good, good) times the number of patterns. The same approach works for 3: here there are $6=\binom 42$ patterns, each with probability $\frac{10}{12}\times\frac{9}{11}\times\frac{2}{10}\times\frac{1}{9}$. This works because the first "good" (wherever you put it) always has $10$ on top, the second "good" always has $9$, and so on.
