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Another Bayes' formula problem

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    There are three identical cards that differ only in color. Both sides of one are black, both sides of the second are red, and one side of the third card is black and its other side is red. These cards are mixed up and one of them is selected at random. If the upper side of this card is red, what is the probability that its other side is black?

    2. Relevant equations
    A is upper red, Y is lower black
    P(A) = 3/6
    P(Y) = 3/6

    I did 3/6 because I considered 3 sides are red and 3 are black, but this keeps nagging at me like I am missing something.

    3. The attempt at a solution
    P(Y|A) = P(A|Y)P(Y) / [P(A|Y)P(Y) + P(A|Yc)P(Yc)] =
    1/3 * 3/6 / [1/3 * 3/6 + 1/3 * 3/6] = .5

    I believe this is right, but it almost seems to be too easy to get to this without using Bayes' formula at all, but that might just be a coincidence due to the easy numbers being manipulated. Thanks for any help
     
  2. jcsd
  3. Dec 3, 2008 #2

    marcusl

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    I'm not so sure. You give P(A|Y) = 1/3, but: if the front is red, the back could (a priori) be either black or red. I'd say P(A|Y) = 1/2.

    Actually I'm going to change notation to B and R 'cause I can't keep A and Y straight.
    I learned Bayes' theorem as:

    p(B|R) = p(R|B) p(B) / P(R)
    where p(B|R) is probability that back is black given that front is red.

    The total prob of finding black or red is the same, p(B) = p(R) = 1/2,
    and p(R|B) = 1/2 by the reasoning given above.
    So p(B|R) = 1/2

    As you point out, we didn't need Bayes' theorem to get that. So you got the right answer, maybe the wrong way.
     
    Last edited: Dec 3, 2008
  4. Dec 4, 2008 #3

    Office_Shredder

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    This is wrong If you pick a card, and it has red on one side, 2/3s of the time it's going to be the card with two red sides, and 1/3 of the time it's going to be the card with the black side. So the probability of seeing black on the other side given that you've seen red on one side is going to be 1/3.

    Considering the symmetry of A and Y, calculating P(Y|A) from P(A|Y) is a bad plan to start with since you know they're the same. Instead let X be the event you got the card with one color on each side, and R the event that you see red on the side you're looking at. Find

    P(X|Y) = P(Y and X) / P(Y)

    P(Y and X) = 1/6 by considering there are six sides you could see, and exactly one of them is the red side on the multicolored card. P(Y) = 1/2
     
  5. Dec 4, 2008 #4

    marcusl

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    ah, that makes perfect sense. Thank you! I'll apply my earlier comment to myself: right answer but wrong method!
     
  6. Dec 5, 2008 #5
    ok, that helped a lot, thanks
     
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