2. (1984 American High School Mathematics Examination) A box contains 11 balls, labeled $1,2,3, \cdots$,
11. If 6 balls are randomly drawn from the box, the probability that the sum of the numbers on the drawn balls is odd is ( ).
A. $\frac{100}{231}$
B. $\frac{115}{231}$
C. $\frac{1}{2}$
D. $\frac{118}{231}$
E. $\frac{6}{11}$
2. D. Reason: The sum of the labels of the selected balls is odd if and only if the number of balls with odd labels selected is odd (if it is an even number, then their sum of labels must be even). And selecting 6 balls from 11 balls, the total number of selections is $C_{11}^{6}=$ 462.
The number of ways to select one ball with an odd label is $C_{6}^{1} \cdot C_{5}^{5}=6$;
The number of ways to select 3 balls with odd labels is $C_{6}^{3} \cdot C_{5}^{3}=200$;
The number of ways to select 5 balls with odd labels is $C_{6}^{5} \cdot C_{5}^{1}=30$.
Therefore, the probability sought in the question is $\frac{6+200+30}{462}=\frac{118}{231}$.