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(A) The total number of outcomes when rolling a tetrahedral die three times is $4 	imes 4 	imes 4 = 64$.For the quadratic equation $ax^2 + bx + c =

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(A) The total number of outcomes when rolling a tetrahedral die three times is $4 	imes 4 	imes 4 = 64$.For the quadratic equation $ax^2 + bx + c = 0$ to have real roots,the discriminant $D = b^2 - 4ac$ must be greater than or equal to $0$,i.e.,$b^2 \geq 4ac$.We test values for $b \in \{1, 2, 3, 4\}$:$1$. If $b = 1$,$b^2 = 1$. $1 \geq 4ac$ has no solution for $a, c \in \{1, 2, 3, 4\}$.$2$. If $b = 2$,$b^2 = 4$. $4 \geq 4ac \Rightarrow ac \leq 1$. The only solution is $(a, c) = (1, 1)$. ($1$ case)$3$. If $b = 3$,$b^2 = 9$. $9 \geq 4ac \Rightarrow ac \leq 2.25$. Possible pairs $(a, c)$ are $(1, 1), (1, 2), (2, 1)$. ($3$ cases)$4$. If $b = 4$,$b^2 = 16$. $16 \geq 4ac \Rightarrow ac \leq 4$. Possible pairs $(a, c)$ are $(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (3, 1), (4, 1)$. ($8$ cases)Total favorable outcomes = $1 + 3 + 8 = 12$.The probability is $P = \frac{12}{64} = \frac{3}{16}$.Thus,$m = 3$ and $n = 16$. Since $\operatorname{gcd}(3, 16) = 1$,$m + n = 3 + 16 = 19$.

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