What is the probability of selecting three different integers, without replacement, from the numbers $1$ to $15$ so their sum is divided by $3$? 
What is the probability of selecting three different integers, without replacement, from the numbers $1$ to $15$, inclusive, so that their sum can be exactly divided by $3$?

My solution:
Number of ways of selecting $3$ numbers (so that their sum can be exactly divided by $3$) either, 


*

*$3$  numbers that are $0$ modulo $3$

*$3$ numbers that are $1$ modulo $3$

*$3$ numbers that are $2$ modulo $3$

*$3$ numbers such that one is $0$ modulo $3$, one is $1$ modulo $3$ and one is $2$ modulo  $3$


so, there are    $3\cdot {_5\mathbf C}_3 + 5\cdot5\cdot5 = 155$ favorable outcomes, ${_{15}\mathbf C }_3 = 455$
so, the probability is $\frac{155}{455}$
isn't the procedure correct? 

Correct. Here it is a high-tech alternative. Let we consider:
$$q(x,y)=(1+yx)(1+yx^2)(1+yx^3)\cdot\ldots(1+yx^{15}).$$
If we take the coefficient of $y^3$ in the expansion of $q(x,y)$ we get a polynomial of the $x$-variable, that evaluated at $x=1$ gives the number of subsets of $\{1,2,\ldots,15\}$ with three elements:
$$ [y^3](1+y)^{15} = \binom{15}{3}.$$
If we consider the coefficient of $y^3$ in $q(x,y)$ and we evaluate it at $x=1, x=\omega=\exp\left(\frac{2\pi i}{3}\right)$ and $x=\omega^2$, we get that the number of subsets of $\{1,2,\ldots,15\}$ with three elements adding to a multiple of three is given by:
$$ \frac{1}{3}\left[\binom{15}{3}+2[y^3](1+y)^5(1+\omega y)^5(1+\omega^2 y)^5\right]$$
that is:
$$ \frac{1}{3}\left[\binom{15}{3}+2[y^3](1+y^3)^5\right]=\frac{1}{3}\left[\binom{15}{3}+10\right]=\color{red}{155}.$$
