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Difficult Math Question #30 - Probability

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Question #30 - Probability

Post Fri Oct 20, 2006 10:37 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    OA coming after some people answer:

    5 girls and 3 boys are arranged randomly in a row. Find the probability that:

    A) there is one boy on each end.

    B) There is one girl on each end.

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    ajith Legendary Member
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    Post Sun Oct 22, 2006 9:51 pm
    Total no: of ways to arrange them = 8!

    No: of ways in which boys are the end

    we can select 2 boys from 3 in 3 ways ...

    and arrange them in 2 ways

    we can arrange the remaining 6 in 6! ways


    So the total no of ways = 2*3*6!


    the probability = 6*6!/8!= 6/7*8=3/28


    Similarly no of ways in which gals are in the end = 5c2*2*6!
    =20*6!
    The probability= 20*6!/8!=20/56=5/14

    There can be mistakes ...

    rajs.kumar Senior | Next Rank: 100 Posts Default Avatar
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    Post Mon Oct 23, 2006 7:37 am
    I get the same answer.

    1. 3P2 x 6!/8! = 3/28

    2. 5P2 x 6!/8! = 5/14

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Mon Oct 23, 2006 2:25 pm
    OA:

    For the first scenario:
    A) there is one boy on each end.

    The first seat can be filled in 3C1 (3 boys 1 seat) ways = 3
    the last seat can be filled in 2C1 (2 boys 1 seat) ways = 2
    the six seats in the middle can be filled in 6! (1 boy and 5 girls) ways
    Total possible outcome = 8!
    Probability= (3C1 * 2C1 * 6!)/ 8! = 3/28

    For the second scenario:
    A) there is one girl on each end.

    The first seat can be filled in 5C1 (5 girls 1 seat) ways = 5
    the last seat can be filled in 4C1 (2 girls 1 seat) ways = 4
    the six seats in the middle can be filled in 6! (3 boys and 3 girls) ways
    Total possible outcome = 8!
    Probability= (5C1 * 4C1 * 6!)/ 8! = 5/14

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