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Topic: Probabilities always >= 0 and <= 1?
Replies: 20   Last Post: May 9, 2012 12:14 AM

 Messages: [ Previous | Next ]
 exmathematician Posts: 23 From: Tulsa Registered: 4/30/12
Re: Probabilities always >= 0 and <= 1?
Posted: May 8, 2012 10:01 AM

> 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.

Date Subject Author
5/8/12 FFMG
5/8/12 Ross Clement
5/8/12 exmathematician
5/8/12 FFMG
5/8/12 Jussi Piitulainen
5/8/12 FFMG
5/8/12 exmathematician
5/8/12 exmathematician
5/8/12 FFMG
5/8/12 Ross Clement
5/8/12 Jussi Piitulainen