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Conditional probability drawing two balls from an urn.

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data
    An urn contains four colored balls: two orange and two blue. Two balls are selected at random without replacemen, and you are told that at least one of them is orange. What is the probability that the other ball it orange?


    2. Relevant equations
    Now, we have been studying P(A|B) meaning the probability of A given the B has already happened.



    3. The attempt at a solution
    Now, my book ends up with a solution of 1/5. But I have no idea how. I said since you know that one of the balls is definitely orange, then you have 3 balls to choose from. So if I want to do it technically (1C1) / (3C1) = 1/3 (C stands for the combination) . However, it's just a quite logical problem in my opinion.

    So who is correct? Me or the book? Thanks! I would love to know how to get 1/5 if I am wrong.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 13, 2011 #2

    Dick

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    The problem is small enough to just do exhaustively. Suppose the orange balls are o1 and o2 and the blue balls are b1 and b2. There are six possible ways to draw, {o1,o2}, {o1,b1}, {o1,b2}, {o2,b1}, {o2,b2} and {b1,b2}. 4C2. In how many of them is at least one ball orange? In how many of them is the second ball also orange? Now write that in combinatorial notation so you know how to do the same problem with larger numbers.
     
  4. Sep 14, 2011 #3
    Well I'm an idiot. Lol, I don't know why I didn't think of it in this way. I got the combinatorial notation as (1C1)(2C0)/(5C1)

    Thanks for your help!
     
  5. Sep 14, 2011 #4

    Dick

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    That works numerically, but I don't see the logic. Where did 5C1 come from?
     
  6. Sep 14, 2011 #5
    Well, I just said (5c1) is the ways in which we can choose a set from 5 different sets. Another I have been doing problem is to say that I was say wanting the double orange set on the 2nd try. So I changed the combinatoric formula to [(1C0)(4C1)/(5C1)]*(1/4)

    This is read as not choosing the double orange*choosing 1 from the orange-blue combination(remember that we are at the given that there must be at least one orange) divided by the number of combinations possible of choosing 1 from the set of all possibilities. And then I multiply by the probability that the 2nd position get the double orange. This method had been working on a few other problems with that kind of description. It seem to work for me. If I am wrong though, please help me :) Thank you very much for your input!
     
  7. Sep 14, 2011 #6

    Dick

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    There is only one way to choose 2 orange balls, right? That's 2C2. Now how many ways to choose 1 orange ball? It's 2C1*(the number of ways to choose one blue ball), also ok? Can you fill it in from here? You are going to have to ADD the number of ways to pick 2 orange balls to the number of ways to pick 1 orange ball to get the number of ways of picking AT LEAST 1 orange ball.
     
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