Find the probability mass function of a discrete random variable I'm having trouble calculating this function, I've already done something but it is not giving me the correct probabilities given that their addition gives me a number larger than 1.
So here's what the exercise says. 
A fair die is rolled 3 times. Let the discrete random variable X be equal to the number of times in which the result of a die roll is equal to a multiple of 3.
Okay so a die has two multiples of 3 which are 3 and 6. So here's what I've done: 
$X = {0,1,2,3}$
My first result goes as follows: $\frac{C(3,0)*2^3}{6^3}$. I picked $2^3$ because there's two numbers that can be picked in 3 rolls. 
$\Omega = 6^3$ given that the die is rolled 3 times. However when I try to get the next part I get the wrong answer. 
I tried to do this for the case in which 1 multiple of 3 shows up: 
$\frac{C(3,1)2^3}{6^3}$.
If it is any help this is the answer: 
$$f_X(m) = \begin{cases} 
            \frac{1}{27} & \text{if $m = 0$} 
            \\\frac{4}{9} & \text{if $m = 1$} 
            \\\frac{2}{9} & \text{if $m = 2$}
            \\\frac{8}{27} & \text{if $m = 3$}
\end{cases}$$
I'm guessing there's something wrong with how I'm using those combinations, but I'm not sure. If anyone can help I'd appreciate it. In fact if someone could help me figure out what's the general idea to find these functions, I'd greatly appreciate. 
 A: 
I'm guessing there's something wrong with how I'm using those combinations

Just a bit, yes.
The probability of rolling $3$ or $6$ on any particular die, that is a success, is $\tfrac 1 3$, and the probability of failure is $\tfrac 2 3$.
To calculate the probability of rolling $x$ successes and $3-x$ failures you multiply the probabilities of doing so in a certain order by the count of arrangements(ie: permutations of that order).
$$\begin{align}\mathsf P(X{=}x) & = {^3{\rm C}_x} \cdot {(\tfrac 1 3)}^x\cdot{(\tfrac 2 3)}^{3-x} \\[2ex] & = \frac{{^3{\rm C}_x}\,2^{3-x}}{3^3} & =\frac{{^3{\rm C}_x} 2^x 4^{3-x}}{6^3} \\[3ex] \mathsf P(X{=}0) & = \frac{{^3{\rm C}_0} 2^3}{3^3} & \text{etcetera}\end{align}$$
By the way, this is called a binomial distribution.
A: For the case when there is exactly one multiple of 3 amongst the three rolls:
Pick the die which throws a multiple of three in $\binom{3}{1}$ ways. It may take two values. The other two dice can each take $4$ values, one of $\{1,2,4,5\}$. Thus, the probability is $\binom{3}{1} \frac{2\times 4^2}{6^3}$.
Can you get the other bits from here?
