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Probability

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prernamalhotra Rising GMAT Star Default Avatar
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Probability Post Mon Jun 16, 2014 6:14 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

    1) 1/14
    2) 1/7
    3) 2/7
    4) 3/7
    5) 1/2

    Thank you,
    Prerna

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    Post Mon Jun 16, 2014 6:22 am
    prernamalhotra wrote:
    A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

    A) 1/14
    B) 1/7
    C) 2/7
    D) 3/7
    E) 1/2
    One approach is to apply counting methods:

    P(exactly 2 women) = [# of teams with exactly 2 women] / [total # of teams possible]

    # of teams with exactly 2 women
    Take the task of selecting 2 women and 2 men and break it into stages.
    Stage 1: Select 2 women for the team. There are 5 women to choose from, so this can be accomplished in 5C2 ways.
    Stage 2: Select 2 men for the team. There are 3 men to choose from, so this can be accomplished in 3C2 ways.
    By the Fundamental Counting Principle (FCP), the total number of teams with exactly 2 women = (5C2)(3C2) = (10)(3) = 30


    # of teams possible
    There are 8 people altogether and we must choose 4 of them.
    This can be accomplished in 8C4 ways, which equals 70 ways



    P(exactly 2 women) = [30] / [70]
    = 3/7 = D

    Aside: To learn how to calculate combinations (like 5C2) in your head, you can watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=789

    Cheers,
    Brent

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    Post Mon Jun 16, 2014 6:32 am
    6 combinations each with probability (5*4*3*2)/(8*7*6*5) = 3/7

    Post Mon Jun 16, 2014 7:14 am
    Quote:
    A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

    A) 1/14
    B) 1/7
    C) 2/7
    D) 3/7
    E) 1/2
    When a probability problem uses the word exactly, we can apply the following reasoning:
    P(exactly n) = P(one way) * all possible ways.

    Let W = woman and M = man.

    P(one way):
    One way to get 2 women and 2 men is WWMM.
    P(1st person is W) = 5/8. (8 people, 5 of them W.)
    P(2nd person is W) = 4/7. (7 people left, 4 of them W.)
    P(3rd person is M) = 3/6. (6 people left, 3 of them M.)
    P(4th person is M) = 2/5. (5 people left, 2 of them M.)
    Since we want all of these events to happen together, we multiply the fractions:
    P(WWMM) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14.

    All possible ways:
    Any arrangement of WWMM will yield exactly 2 W and 2 M.
    Thus, the result above must be multiplied by the number of ways to ARRANGE the 4 elements WWMM.

    The number of ways to arrange 4 elements = 4!.
    But WWMM includes IDENTICAL elements.
    When an arrangement includes identical elements, we must divide by the number of ways each set of identical elements can be arranged.
    The reason:
    When the identical elements swap places, the arrangement DOESN'T CHANGE.
    Here, we must divide by 2! to account for the two identical W's and by another 2! to account for the two identical M's:
    4!/(2!2!) = 6.

    Thus:
    P(exactly 2 W) = 1/14 * 6 = 3/7.

    The correct answer is D.

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    Post Mon Jun 16, 2014 7:56 am
    The answer will staraight away be this way

    Probability = (favorable outcome) / (Total Outcome)
    = (3C2 x 5C2)/(8C4) = 30/70 = 3/7 ANSWER

    [ 3C2 refers the no. of ways of selecting 2 men out of 3 men and 5C2 refers the number of ways of selecting 2 Women out of 5 Women ]

    [8C4 refers to the total number of ways of selecting 4 out of 8 individuals]

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