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Math Help - More conditional prob

  1. #1
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    More conditional prob

    Suppose there are two bags A and B and a set of colored balls (red and green).

    Partially color-blind people are told to place all the red balls in A and the green balls in B.

    Their ability to distinguish the balls is 80% reliable.

    15% of the balls are red, the rest green. (85%)

    If we randomly choose a ball from bag A, what is the probability of it being red? (or green)?

    -------------

    My work:

    We know that P(R) = .15, P(G) = .85.

    I think we need something like P(R|A) = ?

    Not sure how to proceed

    EDIT: Ah! I think I see. I think that P(A | R) = .8 and P(B|R)=.2, etc... is that right?

    EDIT 2: Hmm.. no, still not sure what P(A) should be though...
    Last edited by scorpion007; May 1st 2010 at 10:18 PM.
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    Suppose there are two bags A and B and a set of colored balls (red and green).

    Partially color-blind people are told to place all the red balls in A and the green balls in B.

    Their ability to distinguish the balls is 80% reliable.

    15% of the balls are red, the rest green. (85%)

    If we randomly choose a ball from bag A, what is the probability of it being red? (or green)?

    -------------

    My work:

    We know that P(R) = .15, P(G) = .85.

    I think we need something like P(R|A) = ?

    Not sure how to proceed

    EDIT: Ah! I think I see. I think that P(A | R) = .8 and P(B|R)=.2, etc... is that right?

    EDIT 2: Hmm.. no, still not sure what P(A) should be though...
    Love these questions:
    My thoughts:
    Let's assume there are 100 balls altogether - 15 red and 85 green. We also need to assume that the partially colour-blind people are told that there are 15 red and 85 green. Thus the instruction to put all the red balls in bag A means that they have to put 15 balls in bag A. (If this assumption is not made....we'll come back to this later!)
    So if they are 80% reliable then they will put 80% of 15 red balls (ie12 red balls) in bag A and therefore 3 green balls to total 15.
    So p(red ball from bag A) = 12/15 = 0.8.

    Now if we don't make the assumption that they know there should be 15 balls put in bag A, they would "see" 80% of the 15 of the red balls as red (ie 12) and also they would see 20% of the 85 green balls as red (ie 17) so they would put 29 balls in bag A. Since only 12 of them were really red, p(red ball from bag A) = 12/29.

    I think I'm liking the second scenario!! Nice question.
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  3. #3
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    Thanks alot! I actually came to the same conclusion independently after I posted

    I believe the second scenario is what we're after -- they're just given a set of balls (without knowing the ratio of red/green).
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    Thanks alot! I actually came to the same conclusion independently after I posted

    I believe the second scenario is what we're after -- in that they may mistake some green balls for red ones and misplace them. I got the same probability as you.
    Cheers.
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