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Homework Help: Probability question?

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Two bags contain marbles.

    Bag A contains 2 white marbles and 2 green marbles.

    Bag B contains 3 white marbles, 5 red marbles, and 4 green marbles.

    If you draw one marble at random from each bag, what is the probability of drawing 1 white marble and 1 green marble?

    I AM STUCK. :|

    Help plz.

    2. Relevant equations

    Independent and dependent events.

    3. The attempt at a solution

    Event A = Drawing 1 white marble

    Event B = drawing 1 green marble

    p(A) = 2 / 4
    p(B) = 4 / 12

    2 / 4 * 4 / 12 = 8 / 48

    For some reason it doesn't seem right to me, too easy. ._.
  2. jcsd
  3. Nov 6, 2012 #2
    I think it would be 5/16+6/16 = 11/16. It doesn't matter which bags the marbles come from as long as you have 1 white and 1 green in your hands.
  4. Nov 6, 2012 #3


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    Either do
    (probability bag B draw is not red)(probability bag A draw is opposite bag B draw)
    consider all cases (there are only 48)
  5. Nov 6, 2012 #4

    Ray Vickson

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    The individual sample points are:
    WG = white from bag A and green from bag B
    GW = green from bag A and white from bag B
    (so the first letter is for bag A and the second is for bag B).

    The event you want is E = {WG,GW}, so P{E} = P(WG) + P(GW).

    How would you compute P(WG) and P(GW)?

  6. Nov 6, 2012 #5


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    You are right- both in what you have done and that you are not done yet. That's the probability of drawing a whit marble from bag A and and green marble from bag B. But now you have to do it the other way around. There are 4 marbles in bag A and 2 of them are green so the probability of drawing a green marble from bag A is 2/4= 1/2. There are 12 marbles in bag B and 3 of the are white so the probability of drawing a white marble from bag B is 3/12= 1/4. The probability of drawing "a green marble from bag A and a white marble from bag B is (1/2)(1/4)= 1/8.
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