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Thread: [SOLVED] Mutually independent

  1. #1
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    [SOLVED] Mutually independent

    Problem:
    Let the three mutually independent events $\displaystyle C_{1},C_{2},C_{3},$ be such that $\displaystyle P(C_{1}) = P(C_{2}) = P(C_{3}) = \frac{1}{4}$.

    Find $\displaystyle P[(C_{1}^c \cap C_{2}^c) \cup C_{3}]$.


    Attempt:

    By the definition of mutually independent events,

    $\displaystyle P(C_{1} \cap C_{2} \cap C_{3}) = P(C_1) P(C_2) P(C_3)$.

    Since they are mutually independent events, they are also pairwise e.g.

    $\displaystyle P(C_1 \cap C_3) = P(C_1) P(C_3)$

    So, by the distribution property i.e. $\displaystyle A \cup (B \cap C) = (A \cup B) \cap (A \cup C) $, then

    $\displaystyle P[(C_{1}^c \cap C_{2}^c) \cup C_{3}] = P[(C_{1}^c \cup C_3 ) \cap (C_2^c \cup C_3)]$

    Then by DeMorgan's Laws,

    $\displaystyle = P [(C_1 \cap C_{3}^c)^c \cap (C_2 \cap C_{3}^c)^c)]$

    Since C1, C2, and C3 are mut. ind., then

    $\displaystyle = P((C_1 \cap C_{3}^c)^c P((C_2 \cap C_{c}^c) $

    But from here I do not know how to continue.

    Thank you for your time.
    Last edited by Paperwings; Feb 2nd 2009 at 12:28 PM.
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  2. #2
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    Recall the that complements of independent events are also independent.
    $\displaystyle P\left[ {\left( {A^c \cap B^c } \right) \cup C} \right] = P\left( {A^c \cap B^c } \right)$$\displaystyle + P(C) - P\left[ {\left( {A^c \cap B^c } \right) \cap C} \right] = P\left( {A^c } \right)P\left( {B^c } \right) + P(C) - P\left( {A^c } \right)P\left( {B^c } \right)P\left( C \right)$
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  3. #3
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    Ah, yes. I completely forgot about that. Thank you.
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