If $m$  is a natural number such that $m\le 5$ , then the probability that the quadratic equation $x^2+mx+\frac{1}{2}+\frac{m}{2}=0$  has real roots is

$x^2+mx+\frac{1}{2}+\frac{m}{2}=0$  has real roots

$\therefore m^2-4^2.1\frac{\left(m+1\right)}{2}\ge 0$

$m^2-2m-2\ge 0$

$m^2-2m\ge 2$

$m^2-2m+1\ge 2+1$

${\left(m-1\right)}^2\ge 1$

$m\in \left\{3,4,5,\cdots \right\}$

But $m\le 5$

All possible value of m is 5

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

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