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Re: Probabilities always >= 0 and <= 1?
Posted:
May 8, 2012 10:01 AM
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> On May 8, 10:35 am, FFMG <spambuc...@myoddweb.com>
> wrote:
> > Hi,
> >
> > I was looking at a a site,
> (http://bionicspirit.com/blog/2012/02/09/howto-build-n
> aive-bayes-class...), basically talking about a Naive
> Bayes Classifier.
> >
> > But in some cases the formula gives me
> probabilities greater than 1.
> > How it is possible?
> >
> > // Total of 18 documents.
> > // * 9 documents out of a total of 18 are spam
> messages
> > // * 3 documents out of those 18 contain the word
> "naughty"
> > // * 3 documents containing the word "naughty"
> have been marked as spam
> > // * 3 documents out of the total contain the
> word "money"
> > // * 3 emails out of those have been marked as
> spam
> >
> > P(spam|naughty,money) = P(money|spam) *
> P(money|spam) * P(spam)
> >
> --------------------------
> > P(naughty) *
> P(money)
> >
> > P(spam|naughty,money) = 3/9 * 3/9 * 9/18 = 2
> > ----------------
> > 3/18 * 3/18
> >
> > But how can a probability be outside of 0 and 1?
> Must I always force the numbers to be between 0 and 1
> and accept that in some cases they will fall outside
> the range?
> >
> > Many thanks for suggestions as to where I might
> have gone wrong.
> >
> > Regards,
> >
> > FFMG
>
> You have made an incorrect independence assumption.
> As both "naughty"
> and "money" are only present in "spam" documents,
> which form half o
> the total number of documents, they are dependent
> variables. But, you
> calculate p(e) as p(money) * p(naughty) which is
> assuming that the
> variables are independent. Hence your problem.
You're joking, right? Making the kind of mistake you describe could not result in getting a probablilty greater than one - he's just multiplyng numbers that are all less than one. Could the mistake be in the calculation
3/9*3/9*9/18=2?
hmmmmm...
FFMG, If your question was sincere, the first thing to address is to learn how to multiply fractions: a/b*c/d= ab/cd.
So 3/9*3/9*9/18=9/81*9/18= 81/(81*18)=1/18.
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