3. (1979 American High School Mathematics Examination) Any ordered pair of integers $(b, c)$ is chosen, where the absolute value of each integer is less than or equal to 5, and each such ordered pair of integers has an equal chance of being chosen. The probability that the equation $x^{2}+$ $b x+c=0$ does not have distinct positive real roots is ( ).
A. $\frac{106}{121}$
B. $\frac{108}{121}$
C. $\frac{110}{121}$
D. $\frac{112}{121}$
E. None of these
3. E. Reason: The following six statements are equivalent:
(1) The equation $x^{2}+b x+c=0$ has a positive root;
(2) The equation $x^{2}+b x+c=0$ has real roots, the smaller of which is positive;
(3) $\sqrt{b^{2}-4 c}$ is a real number and $-b-\sqrt{b^{2}-4 c}>0$;
(4) $0 \leqslant b^{2}-4 c$ and $b<0$;
(5) $0<c \leqslant \frac{b^{2}}{4}$ and $b<0$;
(6) $b=-2$ and $c=1$; or $b=-3$ and $c=1$ or 2; or $b=-4$ and $c=1,2,3$ or 4; or $b=-5$ and $c=1,2,3,4$ or 5.

In (6), it can be seen that unless $b^{2}=4 c$, the roots will not be equal. Thus, removing $(b, c)=(-2,1)$ and $(-4, 4)$ from (6), there remain 10 pairs of ordered integers that can make the equation have distinct positive real roots.
The required probability is $1-\frac{10}{11^{2}}=\frac{111}{121}$.