19. A box contains 11 balls, each marked with a number $1,2, \cdots, 11$. If 6 balls are randomly drawn, the probability that the sum of their numbers is odd is
(A) $\frac{100}{231}$;
(B) $\frac{115}{231}$
(C) $\frac{1}{2}$,
(D) $\frac{118}{231}$;
(E) $\frac{6}{11}$.
D
19. The number of combinations when 6 balls are randomly taken from 11 balls is $\mathrm{C}_{11}^{\circ}$; and the number of combinations where the sum of the numbers on these 6 balls is odd is $\mathrm{C}_{0}^{1} \mathrm{C}_{5}^{5}+\mathrm{C}_{8}^{3} \mathrm{C}_{5}^{3}+\mathrm{C}_{0}^{5} \mathrm{C}_{5}^{1}$, the required probability is
$$
\begin{aligned}
& \frac{C_{8}^{1} C_{8}^{5}+C_{8}^{3} C_{5}^{3}+C_{8}^{8} C_{5}^{1}}{C_{11}^{0}}=\frac{236}{462} \\
= & \frac{118}{231}, \text { i.e., }(D) \cdot
\end{aligned}
$$