<s>[INST] We have $12$ balls numbered with $1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9$ in a bin. 
We have $12$ balls in a bin, numbered from $1$ to $8$ and the other $4$ numbered with the number $9$. We take one of them, write down the number and then put back into the bin again. We do this $3$ times. Which is the probability of obtaining $9$ when we multiply those $3$ numbers obtained?

My try:

There is $\frac{12!}{4!}$ possible cases of extracting $3$ balls from $12$ and obtain a different number on each extraction.
To calculate favorable cases, we see that we need to extract $3$, $1$
and $3$ or $9$, $1$, $1$, up to order of the extractions. We only have
that two options.
How many ways we have to obtain two threes and a one in our extractions? $$\binom{12}{1}\cdot\binom{12}{1}\cdot\binom{12}{1}=36.$$
How many ways we have to obtain one nine and two one in our extractions? $$\binom{12}{4}\cdot\binom{12}{1}\cdot\binom{12}{1}=71316$$
So the probability is:$$P\{\text{obtain a 9 multiplying the results}\}=\frac{36+71316}{12\cdot11\cdot10\cdot\dots\cdot5}$$

I don't know if I did it correctly. Thanks.
 [/INST] I think your methodology is off.  For instance, $\binom{12}{1}\cdot\binom{12}{1}\cdot\binom{12}{1} = 12^3$ (not $36$), is the number of ways to choose any three balls, not to choose two $3$'s and a $1$.

Think about them as $12$ balls: if we pick three of them with replacement, we have $12^3$ possible sequences of balls (where the $9$'s have the same numerical value but are thought of as different balls).  You're correct that the only sequences that work are when we have two $3$'s and a $1$ or when we have two $1$'s and a $9$.  
The total number of ways to draw two $3$'s and a $1$ is $\binom{3}{1} =3$ (we have to choose which of the three balls will be a $1$).
The total number of ways to draw one $9$'s and two $1$'s is $\binom{3}{1} \cdot 4 = 12$ (we have to choose which of the three balls will be a $9$ and then choose which $9$-ball to draw).  
Thus, we have a probability of $$\mathbb{P}(\text{multiplying to }9) =\frac{3 + 12}{12^3} = \frac{5}{576}.$$
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