If a number $x$  is selected from natural numbers 1 to 100, then the probability for $\mathrm{x}+\frac{100}{x}>29$  is
 

$\mathrm{x}+100/\mathrm{x}>29\Rightarrow {\mathrm{x}}^2-29\mathrm{x}+100>0\Rightarrow (\mathrm{x}-4)(\mathrm{x}-25)>0\Rightarrow \mathrm{x}<4 \mathrm{or} \mathrm{x}>25.$
$\text{ Thus }\mathrm{n}\left(\mathrm{S}\right){=}^{100}{\mathrm{C}}_1=100,\mathrm{n}\left(\mathrm{A}\right)=\mathrm{n}\left(\right\{1,2,3,26,27,\dots \dots \dots 100\left\}\right)=78$
$\mathrm{P}\left(\mathrm{A}\right)=\frac{78}{100}=\frac{39}{50}$