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Probability question involving intersections

  1. Feb 20, 2015 #1
    1. The problem statement, all variables and given/known data
    and is in reference to the intersection of a set.
    I know the formula that P(A and B ) = P(B) * P(A|B)

    Now we are given something of the nature of P(A and B and C) I am trying to figure out a formula for this. From my searches on google I only find people giving the formula and I really would like to figure it out for my self.

    Here is my attempt even though it does not appear to align with the correct formula.

    P(A and B and C) = P((A and B) and (C)
    = P(K and C ) ; such that P(A and B) = P(K)
    = P(C) * P(K|C) ( now I need to do ( B|A)
    = P(C) *P(A and B|C)*P(B|A)

    What exactly am I doing wrong? I must have made some sort of logical error.

    Correct answer should be
    P(A)*P(C|A and B) *P(B|A)
     
  2. jcsd
  3. Feb 20, 2015 #2
    If A happens first then B and finally C,
    P(C/K) = P(C and K)/P(K)
    =>P(C and K) = P(K)*P(C/K)
    = P(A and B)*P(C/A and B)

    P(A and B) = P(A)P(B/A) because A happens first and then B takes place.
    Formula is correct. In your formula,B happens first.
     
  4. Feb 20, 2015 #3

    Ray Vickson

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    Note: this last expression is equal to
    [tex] P(A)\, P(C|A \cap B)\, P(B | A) = P(C | A \cap B)\,\underbrace{ P(B|A)\,P(A)}_{=P(A \cap B)} \\
    \hspace{2cm}= P(C | A \cap B) \,P(A \cap B) = P (C \cap A \cap B) [/tex]
    You can go backwards and "undo" ##P(A \cap B \cap C) = P(C \cap A \cap B)## to get the final result you want.
     
    Last edited: Feb 20, 2015
  5. Feb 21, 2015 #4

    haruspex

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    The order of events is immaterial.
     
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