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Probability of randomly selecting three big blue marbles?

  1. Apr 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A bag contains 40 blue marbles, 40 red marbles, 20 green marbles. 60 of the marbles are big. Three marbles are selected without replacement.

    What is the probability of randomly selecting three big blue marbles?

    2. Relevant equations

    p(x,y/theta=p(x)p(y/x)=p(y)p(x/y)

    3. The attempt at a solution

    (40/60)(39/59)(38/58)/(40/100)(39/99)(38/98)
     
  2. jcsd
  3. Apr 21, 2008 #2
    first being big blue: p1 = (40/60)(60/100)
    second: p2 = (39/60)(59/99)
    third: p3 = ...

    all 3 = p1*p2*p3
    ... right?
     
  4. Apr 21, 2008 #3

    exk

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    wouldn't p2 = (39/59)(59/99) since there is 1 less big marble left?
     
  5. Apr 21, 2008 #4
    def (thats what i intended), sorry.
    p2 = (39/59)(59/99)
     
  6. Apr 22, 2008 #5
    after multiplying p1*p2*p3 do I need to divide the sum by another number?
     
  7. Apr 22, 2008 #6

    exk

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    Why would you think that you need to divide by another number?
     
  8. Apr 23, 2008 #7
    Because the answer when p1*p2*p3=.061. yet my worksheet has choices of a. .0125, b. .0240, c. .2323, d. .2400
     
  9. Apr 23, 2008 #8

    exk

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    well on second thought you are write, it can't be the above.

    B=blue, b=big.

    P(B)*P(b) is the probability of the marble being big and blue and it initially = 40/100*60/100 since being big and being blue are independent of each other.

    So I get 40/100*60/100*39/99*59/99*38/98*58/98 for all 3 to be B&b, since they are drawn in succession and without replacement. That = 0.0129, which is close to A?
     
  10. Apr 23, 2008 #9
    Thanks I got it and I really appreciate your help!
     
  11. Apr 23, 2008 #10

    exk

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    so was A the answer? I am just curious.
     
  12. Apr 23, 2008 #11
    yes thanks!
     
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