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Math Help - Independent events

  1. #1
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    Independent events

    I am given that P(A) >0, P(B) >0, P(C) > 0 (strictly greater) and

    P(B)P(C) = \frac{P(B \cap C)P(A \cap B)}{P(A \cap (B \cap C))} + \frac{P(B \cap C^{C})P(A \cap B)P(C)}{P(A \cap (B \cap C^{C})P(C^{C})}

    I am to deduce whether B and C could be independent.

    I am thinking no.

    In order for the right side to be equal to  P(B \cap C) this means at least

    <br />
\frac{P(B \cap C^{C})P(A \cap B)P(C)}{P(A \cap (B \cap C^{C})P(C^{C})} = 0

    Now, P(C)  \neq 0 since it was given that P(C) > 0
    P ( A \cap B) \neq 0, since this would make \frac{P(B \cap C)P(A \cap B)}{P(A \cap (B \cap C))} = 0

    Therefore P(B \cap C^{C}) = 0

    However P(B \cap C^{C}) = 0 implies B and C^{C} are not independent in this case because P(B) > 0 and P(C^{C}) > 0 ( since P(C) > 0)

    B and C^{C} not independent implies B and C aren't independent.

    Is this sound reasoning? I am skeptical because this based of the converse of a true statement.

    Thank you,

    -Jame
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  2. #2
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    Wait

    P(C^{C}) is not necessarily greater than zero. ( i.e. if P(C) = 1)

    Darn, well at least what I said still works if P(C) \neq 1
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  3. #3
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    Edit: I forgot that we were assuming that P(A | B \cap C) \neq P(A|B).

    So the P(C) is not 1 and the P(C^{C}) is not zero.
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  4. #4
    MHF Contributor matheagle's Avatar
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    Did you try a few numbers to see whether there might be a counter example?
    maybe drawing a Venn diagram and throwing in some numbers where P(BC)=P(B)P(C) might shed light one way or another.
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