I have the following problems.
In a box we have $3$ balls of color green, violet, orange, blue (so in total $12$ balls). We pick four of them without replacements, what is the probability that there is no orange ball among the four?
My idea was the following, let us define $$\Omega=\left\{f:\{1,2,3,4\}\rightarrow \{g,g,g,v,v,v,o,o,o,b,b,b\}~s.t.~f ~~\text{is injective}\right\}$$ then $|\Omega|=12\cdot 11\cdot 10\cdot 9$. Now define $\Lambda\subset \Omega$ such that for $(a,b,c,d)\in \Lambda$ we have $a,b,c,d\neq o$.So we only have $9$ elements which we can use, i.e. $|\Lambda|=9\cdot 8\cdot 7\cdot 6$. Therefore $$\Bbb{P}(\Lambda)=\frac{9\cdot 8\cdot 7\cdot 6}{12\cdot 11\cdot 10\cdot 9}=\frac{14}{55}$$
Does this work so or is this completely wrong?
Thanks for your help.
In your edited question, your answer is correct and your analysis is good. However, in my opinion, a better approach to these types of questions is a combinatorics approach, where the probability is computed as
$$\frac{N\text{(umerator)}}{D\text{(enominator)}}.$$
$D$ which represents the total number of ways of selecting $4$ distinct balls, without replacement, from the $(12)$ balls, where order of selection is (arbitrarily) deemed irrelevant is equal to
$$ \binom{12}{4}.$$
With the $(3)$ red balls excluded from selection, $N$ then represents the total number of ways of selecting $4$ distinct balls, without replacement, from the $(9)$ remaining balls. Again, order of selection is regarded as irrelevant, primarily because the numerator and denominator must be computed in a consistent manner. So, $N$ equals
$$ \binom{9}{4}.$$
The (somewhat) alternative approach, that more closely relates to your mathematical calculations is to regard order of selection as important in both the numerator and denominator. Then, you have that
$$N = \frac{9!}{(9-4)!} ~~\text{and}~~ D = \frac{(12)!}{[(12) - 4]!}.$$