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kolmogorov's

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  • neotitecs
    HI if A and B are events in & and P(B) =0 then the conditional probability of A given B(written) P(A/B)= P(ANB)/P(B) N=intersect sign P( /F) is a
    Message 1 of 1 , Oct 25, 2003
      HI
      if A and B are events in & and P(B)>=0 then the conditional
      probability
      of A given B(written) P(A/B)= P(ANB)/P(B)
      N=intersect sign
      P('/F) is a probability,conditional probabilities satisfy the
      kolmogorov's 1)-2)-3) axioms

      1) 0<=P('/F)<=1
      2) P(&/F)=1

      3) if Ei....... ,i=1,2,3.....
      are mutually exclusive events then

      infinite infinite
      p(U Ei/F) = E P(Ei/F)
      i=1 i=1

      U=union set
      E=Sigma(additional set)

      how would you prove 3rd{(3)} then??

      toyanc yazgan
      http://titecs.com
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