Let $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ denote the outcome of three independent rolls of a fair…

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 ________

Solution

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

Asked in: JEE Main 2024 (09 Apr Shift 1)

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