In article <[EMAIL PROTECTED]>,
 [EMAIL PROTECTED] (Eddie Jaye) wrote:

> Bill Jefferys <[EMAIL PROTECTED]> wrote in message 
> news:<[EMAIL PROTECTED]>...
> > In article <[EMAIL PROTECTED]>,
> >  [EMAIL PROTECTED] (Eddie Jaye) wrote:
> > 
> > 
> Hi Bill
> 
> thank you  for your response. You say that as the questions stands
> that it cannot be solved or calculated using the fisrt 3 lines of
> explanation. prehaps i should have made myself clearer. what i mean to
> say what is the probability of A given the evidence of B and C and D
> which are all conditional probalities and dependent on A but which ARE
> NOT mutually exclusive and independent on each other. Thus as i
> mentioned before a student can belong to a series of stastics that are
> grouped together in terms of age, gender, ethnicity and so on. if the
> solution to this problem is already in some form provided by you as
> given below
> 
> P(A|B&C&D)=P(B|A&C&D)P(C|A&D)P(A|D)/[P(B|C&D)P(C|D)],
> 
> i do apologise and then ask if you could tell me where i can find the
> other mathetical formulas you hinted about. and politely ask for a
> brief explaination as to why it cannot be solved soley using the input
> sources given.

I have replied to Eddie off-list. There seems to be some confusion here. 
Eddie wants to derive P(A|B&C&D) from P(A|B), P(A|C) and P(A|D). Now of 
course there is the well-known formula 

   P(B&C&D|A)=P(B|A)P(C|A)P(D|A) 

when B, C and D are independent; but there is not to my knowledge anything 
analogous, involving only P(B|A), P(C|A) and P(D|A), when you switch the 
order. One does have, by an obvious application of Bayes' rule to each term 
in the above formula:

   P(A|B&C&D)=P(A|B)P(A|C)P(A|D)xP(B)P(C)P(D)/(P(B&C&D)P(A)^2),

But this involves additional quantities not mentioned by Eddie. 

If B, C and D are unconditionally independent, one can further simplify this 
by factorizing P(B&C&D). The result is

   P(A|B&C&D)=P(A|B)P(A|C)P(A|D)/P(A)^2.

That's as close as I can get to what you want, Eddie! You need some strong 
assumptions about independence, and you need the unconditional P(A), in 
addition to what you stated originally.

Bill

-- 
Bill Jefferys/Department of Astronomy/University of Texas/Austin, TX 78712
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