Hello, Crimsonfear!

I hope you're familiar with Combinations and Factorials.

1. A Box Contains 11 balls, numbered 1, 2, 3, 4, ..., 11.

If 6 balls are drawn simultaneously at random,

what is the probability that the sum of the numbers on the ball drawn is odd.

A) 100/231 . , B) 115/231 . . C) 1/2 . . D) 118/231 . . E) 6/11

There are: .C(11,6) .= .462possible outcomes.

For six numbers to have an odd sum, there are three cases.

[1] Five even, one odd.

. . .There are: .C(5,5)·C(6,1) .= .1·6 .= .6 ways.

[2] Three even, three odd.

. . .There are: .C(5,3)·C(6,3) .= .10·20 .= .200 ways.

[3] One even, five odd.

. . .There are: .C(5,1)·C(6,5) .= .5·6 .= .30 ways.

Hence, there are: .6 + 200 + 30 .= .236ways to have an odd sum.

Therefore: .P(odd sum) .= .236/462 .= .118/231. . . answer choice (D)