An ordered pair $( b , c )$ of integers, each of which has absolute value less than or equal to five, is chosen at random, with each
such ordered pair having an equal likelihood of being chosen. What is the probability that the equation $x^ 2 + bx + c = 0$ will
not have distinct positive real roots?
$\textbf{(A) }\frac{106}{121}\qquad \textbf{(B) }\frac{108}{121}\qquad \textbf{(C) }\frac{110}{121}\qquad \textbf{(D) }\frac{112}{121}\qquad \textbf{(E) }\text{none of these}$
$\boxed{E}$
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 $b0$. 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 $\frac{111}{121}$