$14 \cdot 14$ Choose an ordered pair of integers $(b, c)$, where each integer has an absolute value less than or equal to 5, and each such ordered pair is equally likely to be chosen. The probability that the equation $x^{2}+b x+c=0$ has no 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.
(30th American High School Mathematics Examination, 1979)
［Solution］The following six propositions 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) $0 \leqslant b^{2}-4 c<b^{2}$ 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.

From (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 $1-\frac{10}{11^{2}}=\frac{111}{121}$. Therefore, the answer is $(E)$.