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

$x\in \left[1,5\right]$

$\in \{1,2,3,4,5,\}\Rightarrow n\left(s\right)=5_{c_{{}_1}}$

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

$\left(x-1\right)\left(x-2\right)>0$

$x\in (-\infty ,1)\cup \left(2,\infty \right)\Rightarrow n\left(A\right)=3_{c_{{}_1}}$

$\therefore P\left(A\right)=\frac{3}{5}$