A box contains $11$ balls, numbered $1, 2, 3, ... 11$. If $6$ balls are drawn simultaneously at random, what is the [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that the sum of the numbers on the balls drawn is odd?
$\mathrm{(A) \ }\frac{100}{231} \qquad \mathrm{(B) \ }\frac{115}{231} \qquad \mathrm{(C) \ } \frac{1}{2} \qquad \mathrm{(D) \ }\frac{118}{231} \qquad \mathrm{(E) \ } \frac{6}{11}$
There are exactly $3$ ways the sum can be odd: exactly $5$ are odd, exactly $3$ are odd, or exactly $1$ is odd. There are exactly $6$ odd numbers in the set. For the first case, there are ${6\choose5}\times{5\choose1}=30$ possibilities. For the second case, there are ${6\choose3}\times{5\choose3}=200$ possibilities. For the third case, there are ${6\choose1}\times{5\choose5}=6$ possibilities. In each case, we're choosing out of the $6$ odd numbers which ones will be picked, and multiplying that by out of the $5$ even numbers, which one of those will be picked so that there are $6$ numbers. In total,there are $236$ desirable possibilities out of a total of ${11\choose6}=462$ possibilities. Thus, the answer is $\frac{236}{462}=\frac{118}{231}, \boxed{\text{D}}$.