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If $\mathrm{x} \in[0,5]$, then what is the probability that $\mathrm{x}^{2}-3 \mathrm{x}+2 \geq 0$ ?
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Verified Answer
The correct answer is:
$\frac{4}{5}$
Let $x^{2}-3 x+2=0$
$\Rightarrow \mathrm{x}=1.2$
$\therefore \mathrm{x}^{2}-3 \mathrm{x}+2 \geq 0$ for $\mathrm{x} \in[0,1] \cup[2,3] \cup[3,4] \cup[4,5] .$
It is given that :
$\mathrm{x} \in[0,1] \cup[1,2] \cup[2,3] \cup[3,4] \cup[4,5]$
$\therefore$ Required probability $=\frac{4}{5}$
$\Rightarrow \mathrm{x}=1.2$
$\therefore \mathrm{x}^{2}-3 \mathrm{x}+2 \geq 0$ for $\mathrm{x} \in[0,1] \cup[2,3] \cup[3,4] \cup[4,5] .$
It is given that :
$\mathrm{x} \in[0,1] \cup[1,2] \cup[2,3] \cup[3,4] \cup[4,5]$
$\therefore$ Required probability $=\frac{4}{5}$
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