In article <[EMAIL PROTECTED]>, [EMAIL PROTECTED] (Eddie Jaye) wrote:
> As a new member to this newsgroup I wonder if there is any body out > there that can help me with a probability problem that is part > bayesian part conventional. The problem being to calculate the > probability of two or more events that are not independent of each > other or mutually exclusive. For example what is the probability of my > passing a degree based on previous student records, in which I > simultaneously belong to several event classes age, ethnicity, > gender etc. all of which have their own statistics. > > In mathematical terms > > There exist the individual probabilities of: > > P(A/B) > > P(A/C) > > P(A/D) > > Thus what is the probability of P(A/B and C and D) ? In general it cannot be calculated from the information given on the first three lines. Using '|' instead of '/' (since '/' is needed for division) one has (for example) P(A|B&C&D)=P(B|A&C&D)P(C|A&D)P(A|D)/[P(B|C&D)P(C|D)], by repeated application of Bayes' rule, assuming that I haven't made a mistake. There are other formulas that give the desired result, but none of them can be derived solely from P(A|B), P(A|C) and P(A|D). Bill -- Bill Jefferys/Department of Astronomy/University of Texas/Austin, TX 78712 Email: replace 'warthog' with 'clyde' | Homepage: http://quasar.as.utexas.edu I report spammers to [EMAIL PROTECTED] Finger for PGP Key: F7 11 FB 82 C6 21 D8 95 2E BD F7 6E 99 89 E1 82 Unlawful to use this email address for unsolicited ads: USC Title 47 Sec 227 . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
