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Math-probability

Discussion in 'DAT Discussions' started by joonkimdds, May 25, 2008.

  1. joonkimdds

    joonkimdds Senior Member
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    There are 3 blue, 4 green, 5 red balls. What is the probability of drawing a blue, a green and a red ball without replacement?

    I thought answer should be 1/22 becuz
    3/12 x 4/11 x 5/10 = 1/22 but that's not it.
    The solution says I should multiply it by 6 but i don't know why.

    Here is a similar question

    Superwoman has 3 blue balloons, 4 green balloons and 5 red balloons. How many possible 3 balloon bouquets can she make?

    Again, i thought it should be 1/22 but no.
    This is actually different from previous one.
    The previous one multiplied by 6, but this time we divide by 6


    Basically I wanna know
    when I should multiply and when I need to divide.
    and also, when I don't have to do any of those two and just leave the answer 1/22.
     
  2. vvvv

    2+ Year Member

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    the first problem. you should find the pr of 3 balls from 12. it's 3C12. then find the pr of 1blue = 1C3, 1 green = 1C4, 1 red= 1C5

    then (1blue * 1 red * 1 green)/ pro of 3C12 = 3/11
     
  3. vvvv

    2+ Year Member

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    for the second problem. I assume that the all the red ball are identical, all the green balls are identical and the same with blue. so the problem ask how many ways to pick 3 ball from 3 different colors. pr= 3C3+3-1 = 3C5 = 10.

    but if the balls are different, then pr= 3P12.

    In my opinion, I think all red balls are identical, same with green and blue.

    Does anyone know when to assume this?
     
  4. Zerconia2921

    Zerconia2921 Bring your A-game!
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    ...
     
  5. joonkimdds

    joonkimdds Senior Member
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    bravo
     
  6. DentalP87

    2+ Year Member

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    hello. so ehre goes my explanations:

    1) you are right on the 3/12 x 4/11 ... = 1/22 part
    but you forgot to times this by the number of ways that these balls can be chosen, ie. you can choose one red-blue-green, blue-red-green... etc

    so, you have to times 1/22 by 3P3 to account for the different ways of choosing the balls

    2) This question has nothing to do with probability, so you shouldn't even have a fraction as an answer.

    My answer would be 12C3, since order here would not matter (ie. blue-green-red is same as blue-red-green) --> but this question is very unclear imo

    hope this helps
     

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