Rao

ps:

[Notice that it is not all that hard to compute it too.

Suppose you are computing P(C|E) and you write it as k*0.33

Now, suppose you also compute P(~C|E) (where ~C means it is not in class C)-- this too will have P(E) in the denominator and so it too will have the same k factor. Suppose it is k*0.44

now, you know that P(C|E) + P(~C|E) will be 1.0.

So k*0.33+k*0.44 = 1

k = 1/0.77

On Wed, Nov 12, 2008 at 2:43 PM, Balzer, Michael J (Mike) <Michael.Balzer@asu.edu> wrote:

Hi Dr. Rao,

Regarding P(E), or, P(D) in NBC, just wanted to clarify – since this is a common factor, then it can be safely ignored, correct? In your example slide (willwait), I think you show this as "k" in the calculation. In the homework, do we just show this as k in the final answer?

Thanks again,

Mike Balzer

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