1. We roll a fair tetrahedral die three times, so $a,b,c \in \{1,2,3,4\}$ independently.

2. Total number of possible ordered triples $(a,b,c)$ is $4^3=64.$

3. For the quadratic $ax^2+bx+c=0$ to have all real roots, its discriminant must be non-negative: $b^2-4ac \ge 0.$

   Since $a\in\{1,2,3,4\}$, we always have $a\ne 0$, so it is indeed a quadratic.

4. We now count the number of triples satisfying $b^2 \ge 4ac.$

Case 1: $b=1$

Then $b^2=1,$ so we need $1\ge 4ac.$ But $a,c\ge 1$, hence $4ac\ge 4$. Impossible.

Number of favorable triples: $0$.

Case 2: $b=2$

Then $b^2=4,$ so we need $4\ge 4ac \implies ac\le 1.$ Since $a,c\ge 1$, this happens only when $(a,c)=(1,1).$

Number of favorable triples: $1$.

Case 3: $b=3$

Then $b^2=9,$ so we need $9\ge 4ac \implies ac\le \frac{9}{4}.$ Thus $ac\le 2.$ Possible ordered pairs $(a,c)$ are: $(1,1), (1,2), (2,1).$

Number of favorable triples: $3$.

Case 4: $b=4$

Then $b^2=16,$ so we need $16\ge 4ac \implies ac\le 4.$ Possible ordered pairs $(a,c)$ with $a,c\in\{1,2,3,4\}$ are:

• $a=1$: $c=1,2,3,4$
• $a=2$: $c=1,2$
• $a=3$: $c=1$
• $a=4$: $c=1$

So total pairs: $4+2+1+1=8.$

Number of favorable triples: $8$.

5. Total favorable triples: $0+1+3+8=12.$

Therefore, required probability is $\frac{12}{64}=\frac{3}{16}.$

So, $m=3,\quad n=16.$ Hence, $m+n=3+16=19.$

Final Answer

$\boxed{19}$