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Homework Help: Urn Probability Question

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data

    An urn contains one white chip and a second chip that is equally likely to be white or black. A chip is drawn at random and returned to the urn. Then a second chip is drawn. What is the probability that a white appears on the second draw given that a white appeared on the first draw.

    2. Relevant equations

    3. The attempt at a solution

    I am trying to use the equation

    P(white2 | white 1) = p(white2 and white 1) / P(white1)

    The back of the book says the answer is 5/6.

    I am fairly sure that P(white1) is (1/2)(2/2) + (1/2)(1/2) or .75

    So that would make p(white2 and white1) to be 5/8 which I can't for the life of me see why.

    Any ideas on how to do this? The examples we work out in class are so elementary and I get bored. The problems assigned for homework as very hard (or at least this seems much much harder).

    Thanks for any help!
  2. jcsd
  3. Jan 17, 2010 #2


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    Can you calculate P(white2 and white1) for yourself, just from the given conditions? Use the same method as you used for P(white1), which is indeed correct.

    Cheers -- sylas
  4. Jan 17, 2010 #3
    Here's what I have. :D

    P(white2 and white1) = (.5)(2/2) + (.5)(1/2)(1/2) = 5/8 which is what I need.

    I didn't see it the first time that the first 1/2 represents the first one HAVING to be white while the second 1/2 is the change for the second marble, assuming the variable marble is black.

    Is this correct thinking?
  5. Jan 17, 2010 #4


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    Yes, I think so. You've now solved the problem.
  6. Jan 18, 2010 #5
    Don't really want to start a whole new topic on this so I'l append this topic.

    "Two fair dice are rolled. What is the probability that the number on the first die was at least as large at 4 given that the sum of the two dice was eight?"

    Here's what I take from it. There are only 8 possibilities for two dice to be eight. (4,4)(5,3),(6,2) and (3,5)(2,6)

    This leaves 3/5 for the chance that the first dice was greater than 4 if the total was 8.

    Of course there are 36 different roles one could get.

    What I am trying to solve is P(1st die in >= 4 | total was eight)

    which is equal to

    p(1st die >=4 AND total was eight) / p(total was eight)

    If I add 3/5 + 5/16 then divide by 5/36 I get an invalid answer. Obviously I am suppose to divide.

    Why before did I add before and this time I am using multiplication?
  7. Jan 18, 2010 #6


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    The probability the first die is 4 or greater and the total is 8 is 3/36. As you noted, there are the three ways to achieve that outcome and there are 36 possible outcomes in total.
  8. Jan 18, 2010 #7


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    I don't understand what you did. Can you elaborate?

    I don't think this is a simple conditional probability problem; it's a Bayes' Theorem problem. Before the first draw, the probability of the second chip being white is 0.5, but after you draw a chip, you have additional information as to what color the second chip may be. For example, if you drew a black chip first, you would now know for certain the second chip is black, and the probability of a white on the second draw would be 0.5, compared to 0.75 on the first draw. In the problem, since the first draw was white, it's now more likely that the second chip is white. The idea is to find this new probability to determine the chance the second chip drawn will be a white chip.
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