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?

Method 1:  The number of ways of selecting a subset of size $k$ from a set with $n$ elements (the number of unordered selections of $k$ elements from a set of $n$ elements) is 
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
There are $$\binom{12}{4}$$ ways to select four of the twelve components.
The number of ways of selecting exactly $k$ of the $2$ defective components and $4 - k$ of the $12 - 2 = 10$ good components is 
$$\binom{2}{k}\binom{10}{4 - k}$$
Therefore, the probability of selecting exactly $k$ defective components is 
$$\Pr(\text{exactly}~k~\text{defective}) = \frac{\dbinom{2}{k}\dbinom{10}{4 - k}}{\dbinom{12}{4}}$$
Hence, 
\begin{align*}
\Pr(\text{no defective components}) & = \frac{\dbinom{2}{0}\dbinom{10}{4}}{\dbinom{12}{4}}\\
\Pr(\text{exactly one defective component}) & = \frac{\dbinom{2}{1}\dbinom{10}{3}}{\dbinom{12}{4}}\\
\Pr(\text{exactly two defective components}) & = \frac{\dbinom{2}{2}\dbinom{10}{2}}{\dbinom{12}{4}}
\end{align*}
Method 2:  You have correctly calculated the first two answers.  There are only three possibilities.  The number of defective components selected is either zero, one, or two.  Therefore, 
$$\Pr(\text{exactly two defective components}) = 1 - \Pr(\text{no defective components}) - \Pr(\text{exactly one defective component})$$
