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Conditional Probability

  1. Nov 28, 2014 #1
    ##P(A|A∩B) = \frac{P(A∩(A∩B))}{P(A∩B)} = \frac{P(A∩B)}{P(A∩B)} = 1##
    So given the the event "A and B" as the sample space, the probability of A occurring is 1.

    ##P(A|A∪B) = \frac{P(A∩(A∪B))}{P(A∪B)} = \frac{P(A)}{P(A∪B)}##
    Those two events are independent if and only if the probability of "A or B" occurring is 1, in which case the conditional probability of A equals the probability of A.

    Is my reasoning correct?
     
  2. jcsd
  3. Nov 28, 2014 #2

    BruceW

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    you mean that A and A∪B are independent if P(A∪B)=1 ? yes, this is a special case of independent events. There is one other special case which can mean that A and A∪B are independent. You are trying to make the last equation go from
    [tex]P(A|A∪B) = \frac{P(A)}{P(A∪B)}[/tex]
    to become:
    [tex]P(A|A∪B) = P(A)[/tex]
    right? and you did this by setting P(A∪B)=1. But there is another way also.
     
  4. Nov 29, 2014 #3

    Stephen Tashi

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    We'd have to consider the case [itex] B \cap A = \emptyset [/itex]

    Which two events? [itex] A [/itex] and [itex] A \cup B [/itex] ?

    Suppose [itex] B = \emptyset [/itex].

    I think your results are interesting and worth perfecting by taking care of the exceptional cases.
     
  5. Nov 30, 2014 #4
    What about ##P(A|B)##? My gut tells me that since A and B intersect, A and B must be dependent. This is wrong, of course, but how do I prove the dependence (or independence) of two intersecting events?
     
  6. Nov 30, 2014 #5

    Stephen Tashi

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    I don't know what your are asking.

    You can't prove the dependence or independence of two events that have a non-empty intersection just from knowing the intersection is non-empty. It's a quantitative question that depends on the numerical values of the probabilities involved. If you want to prove a result, you'll have to add more assumptions - something beyond just knowing that the intersection is non-empty.
     
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