If x is an integer and $x\in [1,5]$  then the probability that $x^2-3x+2>0$

$x^2-3x+2>0$

$\Rightarrow x^2-2x-x+2>0$

$\Rightarrow x(x-2)-1(x-2)>0$

$\Rightarrow (x-1)(x-2)>0$

$x\notin [1,2]\text{\hspace{0.17em}\hspace{0.17em}but\hspace{0.17em}\hspace{0.17em}}x\in [1,5]x\text{\hspace{0.17em}\hspace{0.17em}is an integer }$

$E=\{3,4,5\}\Rightarrow n\left(E\right)=3$

$S=\{1,2,3,4,5\}\Rightarrow n\left(S\right)=5$

$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{5}$