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If the roots of each of the equations $2 x^2+x-1=0,3 x^2-10 x+3=0$ and $6 x^2+11 x-2=0$ corresponds to probabilities of three events of a random experiment, then those events are
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exhaustive
Given quadratic equation
$2 x^2+x-1=0 \Rightarrow 2 x^2+2 x-x-1=0$
$\Rightarrow \quad 2 x(x+1)-1(x+1)=0 \Rightarrow x=\frac{1}{2},-1$
and $3 x^2-10 x+3=0 \Rightarrow 3 x^2-9 x-x+3=0$
$\Rightarrow \quad 3 x(x-3)-1(x-3)=0 \Rightarrow x=\frac{1}{3}, 3$
and $6 x^2+11 x-2=0 \Rightarrow 6 x^2+12 x-x-2=0$
$\Rightarrow 6 x(x+2)-1(x+2)=0 \Rightarrow x=\frac{1}{6},-2$
$\because$ Probability of any event lies in interval $[0,1]$ so according to the question probabilities of events $p=\frac{1}{2}, q=\frac{1}{3}$ and $r=\frac{1}{6}$ (let)
$\because p+q+r=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1$
$\therefore$ The events are exhaustive.
$2 x^2+x-1=0 \Rightarrow 2 x^2+2 x-x-1=0$
$\Rightarrow \quad 2 x(x+1)-1(x+1)=0 \Rightarrow x=\frac{1}{2},-1$
and $3 x^2-10 x+3=0 \Rightarrow 3 x^2-9 x-x+3=0$
$\Rightarrow \quad 3 x(x-3)-1(x-3)=0 \Rightarrow x=\frac{1}{3}, 3$
and $6 x^2+11 x-2=0 \Rightarrow 6 x^2+12 x-x-2=0$
$\Rightarrow 6 x(x+2)-1(x+2)=0 \Rightarrow x=\frac{1}{6},-2$
$\because$ Probability of any event lies in interval $[0,1]$ so according to the question probabilities of events $p=\frac{1}{2}, q=\frac{1}{3}$ and $r=\frac{1}{6}$ (let)
$\because p+q+r=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1$
$\therefore$ The events are exhaustive.
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