Let $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked $1,2,3,4$. If the probability that $a x^2+b x+c=0$ has all real roots is $\frac{m}{n}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to _________.

<p>A quadratic equation $ax^2 + bx + c = 0$ has real roots if and only if its discriminant is non-negative. The discriminant $\Delta$ of the quadratic equation is given by:</p> <p>$\Delta = b^2 - 4ac$</p> <p>For the quadratic equation to have all real roots, the discriminant must be non-negative:</p> <p>$\Delta \geq 0$</p> <p>That means:</p> <p>$b^2 - 4ac \geq 0$</p> <p>Given that $a, b, c$ are the outcomes of rolling a fair tetrahedral die, they can each be one of the numbers 1, 2, 3, or 4. Our task is to determine the probability that this condition holds.</p> <p>We need to analyze the cases where $b^2 \geq 4ac$.</p> <p>Let’s consider all possible values for $a$, $b$, and $c$, and count how many of them satisfy the condition. Since there are 4 choices for each of the variables, there are a total of $4 \times 4 \times 4 = 64$ possible combinations.</p> <p>Now, we count the valid combinations where $b^2 \geq 4ac$:</p> <ul> <li>For $a = 1$: $b^2 \geq 4c$</li> <ul> <li>$$b = 1: 1 \geq 4c \rightarrow \text{(Not possible since } c \ \text{must be } \geq 1 \text{ and not zero)}$$</li> <li>$b = 2: 4 \geq 4c \rightarrow c \leq 1 \rightarrow c = 1$ (1 case)</li> <li>$b = 3: 9 \geq 4c \rightarrow c \leq 2 \rightarrow c = 1 \text{ or } 2$ (2 cases)</li> <li>$b = 4: 16 \geq 4c \rightarrow c \leq 4 \rightarrow c = 1, 2, 3, 4$ (4 cases)</li> </ul> <p>Total for $a = 1 = 1 + 2 + 4 = 7$</p> <li>For $a = 2$: $b^2 \geq 8c$</li> <ul> <li>$b = 1: 1 \geq 8c \rightarrow \text{(Not possible)}$</li> <li>$b = 2: 4 \geq 8c \rightarrow \text{(Not possible)}$</li> <li>$b = 3: 9 \geq 8c \rightarrow c \leq 1$ (1 case)</li> <li>$b = 4: 16 \geq 8c \rightarrow c \leq 2$ (2 cases)</li> </ul> <p>Total for $a = 2 = 1 + 2 = 3$</p> <li>For $a = 3$: $b^2 \geq 12c$</li> <ul> <li>$b = 1: 1 \geq 12c \rightarrow \text{(Not possible)}$</li> <li>$b = 2: 4 \geq 12c \rightarrow \text{(Not possible)}$</li> <li>$b = 3: 9 \geq 12c \rightarrow \text{(Not possible)}$</li> <li>$b = 4: 16 \geq 12c \rightarrow c \leq 1$ (1 case)</li> </ul> <p>Total for $a = 3 = 1$</p> <li>For $a = 4$: $b^2 \geq 16c$</li> <ul> <li>$b = 1: 1 \geq 16c \rightarrow \text{(Not possible)}$</li> <li>$b = 2: 4 \geq 16c \rightarrow \text{(Not possible)}$</li> <li>$b = 3: 9 \geq 16c \rightarrow \text{(Not possible)}$</li> <li>$b = 4: 16 \geq 16c \rightarrow c \leq 1$ (1 case)</li> </ul> <p>Total for $a = 4 = 1$</p> </ul> <p>Adding up all the valid cases:</p> <p>$7 + 3 + 1 + 1 = 12$</p> <p>The total number of valid combinations is 12 out of 64. Thus, the probability is:</p> <p>$\frac{12}{64} = \frac{3}{16}$</p> <p>The value of $\mathrm{m} = 3$ and $\mathrm{n} = 16$. The sum $\mathrm{m} + \mathrm{n} = 3 + 16 = 19$.</p> <p>Hence, the answer is 19.</p>