$14 \cdot 4$ A box contains 11 balls, each labeled with a number from $1, 2, 3, \cdots, 11$. If 6 balls are randomly drawn from the box, what is the probability that the sum of the numbers on the drawn balls is odd?
(A) $\frac{100}{231}$.
(B) $\frac{115}{231}$.
(C) $\frac{1}{2}$.
(D) $\frac{118}{231}$.
(E) $\frac{6}{11}$.
(35th American High School Mathematics Examination, 1984)
[Solution] If and only if the number of selected balls with odd labels is odd, the sum of the labels of the selected balls will be odd (if it is an even number, then their sum of labels must be even). When selecting 6 balls out of 11, the total number of ways to select is $C_{11}^{6}=462$.
The number of ways to select one ball with an odd label is
The number of ways to select three balls with odd labels is
The number of ways to select five balls with odd labels is
$C_{6}^{1} \cdot C_{5}^{5}=6 ;$
So the probability required in the problem is
$C_{6}^{3} \cdot C_{5}^{3}=200$;
$C_{6}^{5} \cdot C_{5}^{1}=30$.
$\frac{6+200+30}{462}=\frac{118}{231}$.

Therefore, the answer is $(D)$.