probability:picking 3 different numbers from 1-3n. its known that their sum is devided by 3. whats the probability that each of them is devided by 3 we pick 3 different numbers from 1,2,3,...3n , which their sum is devided by 3 without remainder,what is the probability that each of the number is devider by 3 without remainder?
What I tried:
I started with checkis what is the probability that the sum will be devided by 3 without any remainder:
the highest sum of any trio is 3n+(3n-1)+(3n-2)=9n-3
the lowest sum is 1+ 2 + 3 =6
so there are 9n-3-6=9n-9 different summaries avaiable.(9n-9)/3 are devided by Without remainder.
if so the probability  1/3n-3.
the probability of picking 3 number that are devided by 3 from 1...3n is (3n/3 pick 3)=(n pick 3)= (n(n-1)(n-2))/3!

Let $a$ be the probability that each number is divisible by $3$, and let $b$ be the probability the sum is divisible by $3$. We want $\frac{a}{b}$. 
To get a non-zero probability, we need $n\ge 3$. There are $\binom{n}{3}$ ways to choose $3$ distinct numbers divisible by $3$. 
Now we need to find the number of ways to choose $3$ numbers so that their sum is divisible by $3$. 
The sum can be divisible by $3$ in four different ways. All of the numbers could be divisible by $3$. Or all the numbers could have remainder $1$ on division by $3$. Or all the numbers could have remainder $2$. Or else there could be one of each kind, as in $(3,16,5)$. 
For the first three ways, there are in each case $\binom{n}{3}$ choices. For the one of each kind, there are $\binom{n}{1}^3$ choices, for a total of $3\binom{n}{3}+\binom{n}{1}^3$. 
Thus the required probability is 
$$\frac{\binom{n}{3}}{3\binom{n}{3}+\binom{n}{1}^3}.$$
