What is the probability that the roots of the equation $x^2+x+n=0$ are real, where $n\in \mathbb{N}$ and $n<4$?

For the roots of $a{x}^2+bx+c=0$ to be real, the discriminant $D\ge 0$.Equation: $x^2+x+n=0$$a=1,b=1,c=n$.
$$D=b^2-4ac=1^2-4(1)(n)=1-4n$$
Condition: $1-4n\ge 0\Rightarrow 1\ge 4n\Rightarrow n\le 1/4$.Given $n\in \mathbb{N}$ (Natural numbers: $1,2,3,\dots$).The possible values for $n$ such that $n<4$ are $\{1,2,3\}$.None of these values $\{1,2,3\}$ satisfy the condition $n\le 1/4$.Therefore, there are zero favorable outcomes.Probability $=0/3=0$.