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Difficult Math Problem #95 - Probability

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #95 - Probability

Post Fri Feb 09, 2007 10:32 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    from diff math problems doc. oa coming when some people answer with explanations..

    From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

    A. 1/10

    B. 4/9

    C. 1/2

    D. 3/5

    E. 2/3

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    banona Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Feb 09, 2007 12:30 pm
    I think answer is (3/5). Why ?

    We have a group of 3 boys and 3 girls, 4 children are to be randomly selected.
    all possible outcomes are of number of combinaison of sets of 4 members from a set of six numbers = 6!/2!*4! = 15

    The assess the probability that an event of equal numbers of boys and girls will be selected, we have to count the number of possible sets that countain 2 boys and two girls;
    there are many ways to do such computation, I would go for the following one using combinaisons and multiplication. Rephrase : We have a number of event sets equal to [( number of sets of two girls from 3)*( number of sets of two boys from 3)] = 3*3 = 9

    Therfore probability is : 9/15 = 3/5

    I hope there I was clear;
    looking forward to reading some feed-backs

    Banona

    A. 1/10

    B. 4/9

    C. 1/2

    D. 3/5

    E. 2/3

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Mon Feb 12, 2007 3:25 pm
    OA:

    Total number of ways of selecting 4 children = 6C4 = 15

    with equal boys and girls. => 2 boys and 2 girls. => 3C2 * 3C2 = 9.

    Hence p = 9/15 = 3/5

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