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Math Help - Probability Help

  1. #1
    Crimsonfear
    Guest

    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

    Please help me out please and thank you
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  2. #2
    Super Member

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    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) .= .16 .= .6 ways.

    [2] Three even, three odd.
    . . .There are: .C(5,3)C(6,3) .= .1020 .= .200 ways.

    [3] One even, five odd.
    . . .There are: .C(5,1)C(6,5) .= .56 .= .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)

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