1. Probability Help

Yeah I need some fast help...

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

2. There are 600 student that are divided into 3 groups.
What is the possibility that Bob, Rick, and James are in the same group?

A).1/27 B).1/9 C).1/8 D).1/6 E).1/3

2. 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) .= .462 possible 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 .= .236 ways to have an odd sum.

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