The Bayesian example table and conditional probabilities seems to have a =
problem.
One case in bottom right is duplicate.
Begin forwarded message:
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From: "Chris J. Myers" myers@ece.utah.edu
Subject: Re: HW 4 question
Date: February 11, 2015 at 3:42:51 PM MST
To: Cody MacDonald icmacd@gmail.com
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Actually, the problem is you have this backwards. The entry is:
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P(cro=3D1 | cI=3D0) =3D P(cI=3D0,cro=3D1) / P(cI=3D0) =3D .21/.52 =3D =
.4 (book says .43 which I believe is a mistake).
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Chris
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On Feb 11, 2015, at 2:50 PM, Cody MacDonald <icmacd@gmail.com =
mailto:icmacd@gmail.com> wrote:
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Hi Chris,
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Conceptually this helps yes, but I'm still having issues back =
calculating to match the notes. or getting my two probabilities to add =
to 1.
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P(cI=3D0 | cro=3D1) =3D P(cI=3D0, cro=3D1) / P(cro=3D1) =3D .21/.35 =3D=
.6 which doesn't equal what is in the notes and doesn't add up to 1 =
with P( cro=3D0 | cI=3D0)
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I didn't have problems anywhere else so I must be missing something.
Any additional thoughts would be great.
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Thank you,
Cody
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On Wed, Feb 11, 2015 at 2:05 PM, Chris J. Myers <myers@ece.utah.edu =
mailto:myers@ece.utah.edu> wrote:
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On Feb 11, 2015, at 2:00 PM, Cody MacDonald <icmacd@gmail.com =
mailto:icmacd@gmail.com> wrote:
Hi Chris,
I've been struggling to back calculate from the example in class. =
Specifically when you are calculating the probability further down the =
bayesian network.
Take cro from the example in class. The probability of cro=3D1 when =
cI is 0 is calculated by adding up probabilities of this case given in =
the chart and then dividing by what? I'm hung up on how to come to a =
number to divide by in this more complicated case.
For the case you mention, you add the probabilities of all cases =
where cro=3D1 and cI=3D0, then you divide by those where cro=3D1.
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This makes use of Baye=E2=80=99s Rule which states:
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P(ci | cro) =3D P(cI,cro) / P(cro)
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So, your case is:
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P(cI=3D0 | cro=3D1) =3D P(ci=3D0, cro=3D1) / P(cro=3D1)
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Does this help?
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Chris
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The Bayesian example table and conditional probabilities seems to have a = problem.
One case in bottom right is duplicate.