Correct answer is : 19
\(a, b, c \in\{1,2,3,4\}\)

Tetrahedral dice
\(a x^{2}+b x+c=0\)
has all real roots
\(\Rightarrow \mathrm{D} \geq 0\)
\(\Rightarrow \mathrm{b}^{2}-4 \mathrm{ac} \geq 0\)
Let \( \mathrm{b}=1 \Rightarrow 1-4 \mathrm{ac} \geq 0\) (Not feasible)
\(\mathrm{b}=2 \Rightarrow 4-4 \mathrm{ac} \geq 0\)
\(1 \geq \mathrm{ac} \Rightarrow \mathrm{a}=1, \mathrm{c}=1\),
\(\mathrm{b}=3 \Rightarrow 9-4 \mathrm{ac} \geq 0\)
\(\frac{9}{4} \geq\) ac
\(\Rightarrow \mathrm{a}=1, \mathrm{c}=1\)
\(\Rightarrow \mathrm{a}=1, \mathrm{c}=2\)
\(\Rightarrow \mathrm{a}=2, \mathrm{c}=1\)
\(\mathrm{b}=4 \Rightarrow 16-4 \mathrm{ac} \geq 0\)
\(4 \geq\) ac
\(\Rightarrow \mathrm{a}=1, \mathrm{c}=1\)
\(\Rightarrow \mathrm{a}=1, \mathrm{c}=2 \quad \Rightarrow \mathrm{a}=2, \mathrm{c}=1\)
\(\Rightarrow \mathrm{a}=1, \mathrm{c}=3 \quad \Rightarrow \mathrm{a}=3, \mathrm{c}=1\)
\(\Rightarrow \mathrm{a}=1, \mathrm{c}=4 \quad \Rightarrow \mathrm{a}=4, \mathrm{c}=1\)
\(\Rightarrow \mathrm{a}=2, \mathrm{c}=2\)
Probability \(=\frac{12}{(4)(4)(4)}=\frac{3}{16}=\frac{\mathrm{m}}{\mathrm{m}}\)
\(\mathrm{m}+\mathrm{n}=19\)