<PRE><B>Question 356933</B>
Breaking down the possibilities:
0 odd, 6 even = sum is even
1 odd, 5 even = sum is odd
2 odd, 4 even = sum is even
3 odd, 3 even = sum is odd
4 odd, 2 even = sum is even
5 odd, 1 even = sum is odd
6 odd, 0 even = sum is even
We only need to consider the odd cases given above
i) 1 odd, 5 even.  The number of ways of selecting 1 odd and 5 even is 
{{{C(6,1)*C(5,5) = 6}}}
ii)3 odd, 3 even.  The number of ways of selecting 3 odd and 3 even is 
{{{C(6,3)*C(5,3) = 200}}}
iii)5 odd, 1 even.  The number of ways of selecting 5 odd and 1 even is 
{{{C(6,5)*C(5,1) = 30}}}.
Now the total number of ways of selecting 6 out of 11 balls is {{{C(11,6) = 462}}}.  Therefore the probability that the sum of the numbers on the balls drawn is odd is {{{(6+200+30)/462 = 236/462 = 118/231}}}.  (Since the 3 cases above are mutually exclusive.)


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