I only know that if it is uniformly distributed the formula for probability function will be $\frac{1}{b-a}$ but in my question I have a uniform distribution on $\{-1,0,1\}$.
For example if $X$ has uniform distribution on $(a,b)$ then it will be
$F(x) \begin{cases} 0 & x < a \\ \frac{x-a}{b-a} & a \le x \le b \\ 1 & x > b \end{cases}$
How do I find uniform distribution on $\{-1,0,1 \}$ ?
Isn't $\{-1,0,1\}$ a set with three elements? In that case we have $P[-1]=P[0]=P[1]=1/3$ and zero otherwise. The distribution function will be $$ F(x) \begin{cases} 0 & x < -1 \\ 1/3 & -1 \le x < 0 \\ 2/3 & 0 \le x < 1 \\ 1 & x \ge 1 \end{cases} $$