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REALLY Simple Probability

  1. Oct 6, 2006 #1
    I know this might be really simple, but I get easily confused.

    CASE 1.

    I have a bag with eight marbles, one of them is green and the remaining seven are red.

    If you have to pick a random marble without looking, there is one in eight chances (12.5%) of lifting the green one.

    Is this correct?

    CASE 2.

    I have three identical bags (like the one in case 1).

    What are the chances of picking the three green ones? 3 in 24? (25%) this is how i though it would work out, but logically I predicted that it would have been more difficult to pick the three out of all the 24 (From three different bags), so I thought I was doing something wrong.

    All help will be greatly appreciated.
     
  2. jcsd
  3. Oct 6, 2006 #2
    Hi There,

    In terms of CASE 1:

    P(green)=1/8

    In Terms of CASE 2:

    E1=result of the first bag
    E2=result of the second bag
    E3=result of the third bag

    P(E1nE2nE3)=(1/8)*(1/8)*(1/8)=1/512

    This should be correct, hope it helps

    regards Steven
     
    Last edited: Oct 6, 2006
  4. Oct 6, 2006 #3
    Thank you!! That's what it was!!

    That does answer my doubt.
     
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