Probability that the sum 3 integers is divisible by 3 
I found the sample space, but i was having difficulty finding the favorable outcomes. 

But how do i find numbers that add up to be divisible by 3?

Al integers are of the form $m=3i_m + k_m$ where $i_m$ is an integer and $k_m$ is either $0,1$ or $2$.  As the the $3n$ numbers are consecutive the number of integers where $k_m$ is $0, 1$ or $2$ are equal.
The ways in which $a+b+c$ is divisible by three is that: i) all three have $k_a = k_b = k_c$ or that ii) $\{k_a,k_b,k_c\} = \{0,1,2\}$.  (If $k_i = k_j=0,1,2$ then $k_i + k_j = 0,2,4$ and for $0,2,4 + k_l$ to be divisible by $3$ then $k_l = 0, 1, 2 = k+i=k_j$.  So either they are all the same, or all different.)
The probability of i) $1*\frac{n-1}{3n-1}*\frac{n-2}{3n-2}$ (because $k_a$ may be anything.  The probability of $k_b = k_a$ is $\frac{n-1}{3n-1}$ because there are $n-1$ numbers with the $k$ value of $k_a$ and there are $3n-1$ numbers left.  And so on.)
The probability of ii) is $\frac{n}{3n}\frac{n}{3n-1}\frac{n}{3n-2}*3!$.  There maybe be other ways to calculate this but $\frac{n}{3n}$ is the probability $k_a = 0$ and $\frac{n}{3n-1}$ is the probability $k_b = 1$ given that $k_a = 0$ and $\frac{n}{3n-2}$ is the probability that $k_c =2$ given that $k_a=0$ and $k_b= 1$ and $3!$ is the number of ways we can order $\{k_a,k_b,k_c\}$ to be equal to $\{0,1,2\}$.
As the events are distinct/independant the probability is:$1*\frac{n-1}{3n-1}*\frac{n-2}{3n-2} + \frac{n}{3n}\frac{n}{3n-1}\frac{n}{3n-2}*3!=\frac{(n-1)(n-2)}{(3n-1)(3n-2)} + \frac{2n^2}{(3n-1)(3n-2)}=\frac{3n^2 - 3n +2}{9n^2 - 9n + 2}= \frac{3n^2 - 3n +\frac 23}{9n^2 - 9n + 2}+ \frac 43*\frac1{(3n-1)(3n-2)}=\frac 13 + \frac{4}{3(3n-1)(3n-2)}=1 - \frac{6n(n-1)}{(3n-1)(3n-2)}$ .
For $n=1$ then probability is $1$.  And as $n$ increases the probability decreases and gets closer to $1/3$.
