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Probability of independent events

  1. Jun 29, 2014 #1
    i have drawn the venn diagram for the question... i am not sure it's correct or not. the question states both are independent events. i dont understand the working in part b ... for me , my ans would be p(A) +p(B).. correct me if i am wrong. thanks in advance.
     

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  3. Jun 29, 2014 #2

    Simon Bridge

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    Set A should be the set of all products with defect A
    Similarly for set B.
    Now can you see why the sets intersect?

    What does that look like on a venn diagram? (hint: not like you did).

    Independent events are those whose probability is not affected by the occurance of the other.
    i.e. where P(A|B)=P(A), P(B|A)=P(B)

    Since: P(A|B)=P(AnB)/P(B)
    This means that for independent events, P(AnB)=P(A)P(B).

    (using AnB = "A intersection B")
     
  4. Jun 29, 2014 #3
    the venn diagram look like this?
     

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  5. Jun 29, 2014 #4

    micromass

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    Correct. Do you see now why your answer was wrong? You essentially did the following:

    [tex]\mathbb{P}(A\cup B) = \mathbb{P}(A) + \mathbb{P}(B)[/tex]

    which is only allowed if ##A## and ##B## do not intersect. But ##A## and ##B## do intersect here, so you cannot apply this formula!
     
  6. Jun 29, 2014 #5

    FactChecker

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    To be a little more specific about independent events -- If events A and B are independent, the probability of event A happening is the same whether B occurs or not. And the converse is true. One way to show that in a Venn diagram is to draw A as a horizontal slice of the whole diagram, and B as a vertical slice of the whole diagram. That way the percent of A within B is the same as the percent of A within the entire diagram (and vice versa). The first diagram in the attachment shows independent events and the second diagram shows dependent events venn_diagram_independent_events.png
     
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