smc

There are $10$ cases where the roots are real, positive and distinct. The roots to the quadratic equation are $\frac{-b\pm \sqrt{b^2-4c}}{2}$. In order for $-b-\sqrt{b^2-4c}>0$ we certainly need $b<0$. Additionally, for the roots to be both real and distinct we need $b^2-4c>0$. And lastly, for both roots to be positive we need $-b>\sqrt{b^2-4c}$. Combining these inequalities we determine that $\frac{b^2}{4}>c>0$ and $b<0$. This leaves just $10$ cases $\{(-5,1),(-5,2)(-5,3),(-5,4),(-5,5),(-4,1),(-4,2),(-4,3),(-3,1),(-3,2)\}$ which can easily be checked by hand. The answer is $\frac{111}{121}$ $\boxed{\text{none of these}}$