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Probability again

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data
    A bag contains 3 black marbles, 2 green marbles and 1 white marble. Two marbles are drawn at random from the bag WITH replacement. Find the probabilities that the two marbles are

    (a)black
    (b)of different colours


    2. Relevant equations



    3. The attempt at a solution

    (a)P(both black) = 3/6 x 3/6 = 1/4

    (b) I have two attempts for this.

    Attempt A
    P(of different colours) = P(B,G) + P(B,W) + P(G,W) + P(G,B) + P(W,B) + P(W,G)
    = (3/6)(2/6) + (3/6)(1/6) + (2/6)(1/6) + (2/6)(3/6) + (1/6)(3/6) + (1/6)(2/6)
    = 22/36

    Attempt B
    P(of different colours) = 1 - P(both black) - P(both green) - P(both white)
    = 1 - (3/6)(3/6) - (2/6)(2/6) - (1/6)(1/6)
    = 22/36

    Any problems for any of the attempt above?
     
  2. jcsd
  3. Apr 15, 2012 #2

    tiny-tim

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    hi kenny1999! :wink:

    looks fine :smile:

    (B is probably neater than A)
     
  4. Apr 15, 2012 #3
    but are they both correct?
     
  5. Apr 15, 2012 #4

    tiny-tim

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    yes of course!

    (didn't i say so? :smile:)
     
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