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Depdendent, Independent, and Complementary Events

  1. Nov 23, 2014 #1
    For two events ##A## and ##B##, where ##A∩B ≠ ∅##, is it possible to deduce from a venn diagram whether or not those two events are dependent? Or is such information unattainable from a mere venn diagram? Also, if we define ##C## as the complement of the union of ##A## and ##B##, why must ##C## be dependent on ##A'##? Why is ##P(C)## not equal to ##P(C|A')##?
     
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  3. Nov 23, 2014 #2

    Stephen Tashi

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    It isn't possible to determine ithe independence from the usual sort of Venn diagram because you need quantitative information. If probability is represented on a scale proportional to area, you need to determine if the ratio of the area of [itex]A [/itex] to the total area is the same as the ratio of [itex] A \cap B [/itex] to the area of [itex] B [/itex].

    What are you claiming? - "not always equal to" or "never equal to" ?
     
  4. Nov 23, 2014 #3

    statdad

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    If the Venn diagram is labeled - yes, you can determine independent or not - but with the same dull calculations you'd do if you were simply provided the numerical information itself.

    Are you asking why, if [itex] C = A \cup B [/itex], that [itex] C' = A' \cap B' [/itex] ? If so, you can convince (not form a proof) yourself of why that is by looking at your hypothetical Venn Diagram. The set C consists of the entire region inside the regions that represent A and B: the complement of C is the entire region inside the diagram but outside the two regions for A and B: that region outside is [itex] A' \cap B' [/itex].

    If that is what you meant - why did you pick on the complement of A alone?
     
  5. Nov 24, 2014 #4
    Never equal to.
     
  6. Nov 24, 2014 #5
    What I was asking was: why is the probability of event ##C## different from the probability of event ##C## given ##A'## (or ##B'## really). It does make sense to me; intuitively, since ##P(C|A') = \frac{P(C∩A')}{P(A')}##, and the intersection of ##C## and ##A'## is ##C##, it is clear that ##P(C)## and ##P(C|A')## differ by a factor of ##P(A')##.
    But using this same logic on a venn diagram, can't we just as well come to the wrong conclusion that whenever there is an intersection between two events ##A## and ##B##, ##A## must always be dependent on ##B## since the occurrence of ##B## implies the occurrence of the intersection, which is also a part of ##A##?
     
  7. Nov 24, 2014 #6

    Stephen Tashi

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    [itex] C [/itex] might be the null set.
     
  8. Dec 1, 2014 #7

    FactChecker

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    One way to visualize independent events in venn diagrams is shown in these two examples. The first shows 2 independent events. One is the horizontal blue area and the other is the vertical pink area. For independent events, the fraction of the horizontal area within the vertical area is the same as its fraction within the entire universal square. The second diagram shows dependent events. Most venn diagrams are just shown as blobs where the areas are not useful for drawing any conclusions.
    venn_diagram_independent_events.png

    For C to be independent of A', it must have the same probability with or without A'. But that is the same as saying it has the same probability without A or with A. We know that P(C|A) = 0, so C and A' can not be independent.
     
    Last edited: Dec 1, 2014
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