There are 3 shooters with probability of taking down the target $$ {{4}\over{5}}, {{3}\over{4}}, {{2}\over{3}}$$ When they shot simultaneously $2$ of them takes down the target. What is the probability that third shooter $\textbf{doesn't}$ take down the target.
I am thinking this way. 3 cases
1) $1$, $2$ took down ${4\over5} * {3\over4}$ for this to happend and also * ${1\over3}$ for third to not take down
2) $2$, $3$ $\rightarrow$ ${{3}\over{4}}*{{2}\over{3}}* {{1}\over{5}}$
3) $1$, $3$ $\rightarrow$ ${{4}\over{5}}*{{2}\over{3}}* {{1}\over{4}}$
the final result will be the sum of all those cases $13\over30$
but the answer isn't that :/
what am I doing wrong?
You did calculate the probability that you ask for in the title of the question: the probability that two hunters take down the target and the third one misses. However, in the text of the question you ask a different question:
This question is ambiguous in its formulation. I'll answer two possible interpretations of it.
In one interpretation, if we take seriously the fact that the first part is in the past tense and the second part isn't, we can interpret this as a temporal succession in which two specific hunters have shot down the target and the third one then also tries to do so. In this case, the question is ill-posed, as we don't know which two hunters shot first, or how they were selected. Assuming that they were uniformly randomly selected, the probability that the third shooter misses the target is
\begin{eqnarray*} P(\text{$S_3$ misses}\mid\text{$S_1$ and $S_2$ hit}) &=& \frac{P(\text{$S_3$ misses}\land\text{$S_1$ and $S_2$ hit})}{P(\text{$S_1$ and $S_2$ hit})} \\ &=& \frac{\sum_iP(\text{$S_3$ misses}\land\text{$S_1$ and $S_2$ hit}\mid S_3=H_i)}{\sum_iP(\text{$S_1$ and $S_2$ hit}\mid S_3=H_i)} \\ &=& \frac{\frac34\cdot\frac23\cdot\left(1-\frac45\right)+\frac23\cdot\frac45\cdot\left(1-\frac34\right)+\frac45\cdot\frac34\cdot\left(1-\frac23\right)}{\frac34\cdot\frac23+\frac23\cdot\frac45+\frac45\cdot\frac34} \\ &=& \frac{\frac{13}{30}}{\frac{49}{30}} \\ &=& \frac{13}{49}\;, \end{eqnarray*}
where $H_i$ is the $i$-th hunter in the order in which you specified the hitting probabilities and $S_i$ is the $i$-th hunter in the order in which they shot.
In another interpretation, ignoring the fact that the first part is in the past tense and the second part isn't, we can take the question to ask for the conditional probability of one shooter missing given that at least two shooters hit. This is
\begin{eqnarray*} P(\text{one shooter misses}\mid\text{at least two shooters hit}) &=& \frac{P(\text{one shooter misses}\land\text{at least two shooters hit}) }{P(\text{at least two shooters hit})} \\ &=& \frac{P(\text{exactly two shooters hit}) }{P(\text{at least two shooters hit})} \\ &=& \frac{\frac{13}{30}}{\frac{13}{30}+\frac45\cdot\frac34\cdot\frac23} \\ &=& \frac{13}{25}\;. \end{eqnarray*}