If the roots of each of the equations 2x^2+x- | vedclass

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(B) For the equation $2x^2+x-1=0$, we have $2x^2+2x-x-1=0 \Rightarrow 2x(x+1)-1(x+1)=0$, so $x = \frac{1}{2}, -1$. Since probability $P \in [0, 1]$, w

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exhaustive

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(B) For the equation $2x^2+x-1=0$, we have $2x^2+2x-x-1=0 \Rightarrow 2x(x+1)-1(x+1)=0$, so $x = \frac{1}{2}, -1$. Since probability $P \in [0, 1]$, we take $P_1 = \frac{1}{2}$.For the equation $3x^2-10x+3=0$, we have $3x^2-9x-x+3=0 \Rightarrow 3x(x-3)-1(x-3)=0$, so $x = \frac{1}{3}, 3$. Since probability $P \in [0, 1]$, we take $P_2 = \frac{1}{3}$.For the equation $6x^2+11x-2=0$, we have $6x^2+12x-x-2=0 \Rightarrow 6x(x+2)-1(x+2)=0$, so $x = \frac{1}{6}, -2$. Since probability $P \in [0, 1]$, we take $P_3 = \frac{1}{6}$.Sum of probabilities $P_1 + P_2 + P_3 = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1$.Since the sum of the probabilities of the events is $1$, the events are exhaustive.

If the roots of each of the equations $2x^2+x-1=0$, $3x^2-10x+3=0$, and $6x^2+11x-2=0$ correspond to probabilities of three events of a random experiment, then those events are

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