What is the probability to get a sum of 9 exactly once when tossing 2 die in 3 throws? So I am throwing 2 die at the same time, the probability for me to get a sum of 9 is $\frac{1}{9}$. So when I throw them 3 times, the chances of 9 occuring once should be $\frac{1}{3}$. Then $\frac{1}{9}$x$\frac{1}{3}$=$\frac{1}{27}$ 

Is this right? 
Also how could I calculate it if the sum of 9 occurs at least once in three throws?

I don't know why but I feel that the answer is wrong but I am not sure with what I'm doing wrong.
To begin, we calculate the probability that within a throw of a pair of dice we get the sum of nine.  It helps to consider the dice as being different colors.  We have the thirty-six equally likely results for pairs of dice being $\{(\color{red}{1},\color{blue}{1}),(\color{red}{1},\color{blue}{2}),(\color{red}{1},\color{blue}{3}),(\color{red}{1},\color{blue}{4}),(\color{red}{1},\color{blue}{5}),(\color{red}{1},\color{blue}{6}),(\color{red}{2},\color{blue}{1}),\dots,(\color{red}{6},\color{blue}{6})\}$

(Note: the outcome of $(\color{red}{1},\color{blue}{2})$ is considered to be different than the outcome of $(\color{red}{2},\color{blue}{1})$)

Of these thirty-six equally likely results, exactly four of them yield a sum of $9$, namely $(\color{red}{3},\color{blue}{6}),(\color{red}{4},\color{blue}{5}),(\color{red}{5},\color{blue}{4})$ and $(\color{red}{6},\color{blue}{3})$.  Taking the ratio of the number of favorable outcomes to the total number of possible outcomes (in an equiprobable sample space) gives the probability $\frac{4}{36}$ which simplifies to $\frac{1}{9}$.  You calculated this correctly.



The question you asked was for calculating the probability that at least once in three independent attempts you roll a sum of nine.  (This is distinctly different than the sum being nine on the first and only the first attempt like so many other answers sadly calculated)

To calculate this, you may approach several different ways.

Method 1: Split into cases

We could have the sum be nine at least once in the following seven ways:


The first sum is nine, the second sum is not nine, the third sum is not nine
The first sum is not nine, the second sum is nine, the third sum is not nine
The first sum is not nine, the second sum is not nine, the third sum is nine
The first sum is nine, the second sum is nine, the third sum is not nine
$\vdots$
The first sum is nine, the second sum is nine, and the third sum is nine


Each of these cases are disjoint, and so their probabilities may be added.  Each can be calculated by multiplication principle as they are independent events.  The probability for the first for example being $\frac{1}{9}\times\frac{8}{9}\times\frac{8}{9}$ while the probability for the second is $\frac{8}{9}\times\frac{1}{9}\times\frac{8}{9}$ and the probability for the fourth is $\frac{1}{9}\times\frac{1}{9}\times\frac{8}{9}$, etc...

(The error in several other answers was in only considering the first of these cases and none of the others)

This is horribly tedious but will arrive at a final correct answer if you are careful enough.



Method 2: Split into cases in a smarter way

We could have the sum be nine at least once in the following three ways:


Exactly one of the throws summed to nine
Exactly two of the throws summed to nine
Exactly three of the throws summed to nine


To calculate these, we can apply the binomial distribution.  Exactly $k$ successes in $n$ attempts where each attempt is independent and has success with probability $p$ will occur with probability $\binom{n}{k}p^k(1-p)^{n-k}$

For this, your first would occur then with probability $\binom{3}{1}(\frac{1}{9})^1(1-\frac{1}{9})^{3-1}$, etc...  We can then sum these as well to arrive at a final answer.  Still a bit tedious for my liking.



Method 3: Break into cases even smarter and use complements

The following two outcomes are possible
 - At least one of the throws summed to nine
 - None of the throws summed to nine

As these are disjoint and cover all cases, their probability must sum to $1$, so to calculate that at least one of the throws summed to nine, it suffices to calculate the probability that none of the throws summed to nine and subtract it from $1$.

None of them summed to nine with probability $(\frac{8}{9})^3$, so the answer to your original question is $1-(\frac{8}{9})^3$



Method 4: Inclusion-exclusion

Letting $A,B,C$ represent the events that the first, second, and third throws ended with a sum of nine respectively, you are asking to calculate $Pr(A\cup B\cup C)$

Expanding via inclusion-exclusion we get 

$Pr(A\cup B\cup C)=Pr(A)+Pr(B)+Pr(C)-Pr(A\cap B)-Pr(A\cap C)-Pr(B\cap C)+Pr(A\cap B\cap C)$

Then using that the events are independent, we expand a bit further to get a final answer of

$\frac{1}{9}+\frac{1}{9}+\frac{1}{9}-\frac{1}{9}\times\frac{1}{9}-\frac{1}{9}\times\frac{1}{9}-\frac{1}{9}\times\frac{1}{9}+\frac{1}{9}\times\frac{1}{9}\times\frac{1}{9}$

(The error in another answer was in only using the first few terms of this expansion and ignoring all of the other still relevant terms)
So the possible combinations of die rolls that will sum to 9 are 6,3 3,6 4,5 and 5,4. The probability of getting these exact combinations is clearly $\frac{1}{36} $ each. So the probability of getting 9 on a given throw of 2 dies is 1/9.
Obviously, the probability of not getting 9 is $ \frac {8}{9} $. So in three throws, you want 9, not 9 and not 9 in any given order. The probability of this is the product of all 3 probabilities times 3 ( so 9 miss miss , miss 9 miss and miss miss 9)

$ 3x \frac{1}{9} x \frac{8}{9} x \frac{8}{9} = \frac{64}{243} = 0.2634 $

This method can easily be applied to any other number by calculating its possible combinations.