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Math Help - Probability of the union of independent events proof (induction argument)

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    Probability of the union of independent events proof (induction argument)

    I am trying to prove the following:

    If E_1, E_2, ... , E_n are mutually independent events, then
    <br />
P(E_1 \cup E_2 \cup ... \cup E_n) = 1 - \Pi^{n}_{i=1} [1 - P(E_i)|<br />

    This is true for n=2

    <br />
P(E_1 \cup E_2) = 1 - P(E_1^{C} \cap E_2^{C}) by DeMorgan's Law

    and since E_1 and E_2 are independent, E_1^{C} and E_2^{C} are independent

    Therefore  P(E_1 \cup E_2) = 1 - P(E_1^{C} \cap E_2^{C}) = 1-P(E_1^{C})P(E_2^{C})

    Assume true for n
    P(E_1 \cup E_2 \cup ... \cup E_n) = 1 - P(E_1^{C} \cap E_2^{C} \cap ... \cap E_n^C) = 1-P(E_1^{C})*P(E_2^{C})*...*P(E_n^{C})

    Then P(E_1 \cup E_2 \cup ... \cup E_n \cup E_{n+1}) = P((E_1 \cup E_2 \cup ... \cup E_n) \cup E_{n+1})<br />

    Thus by the theorem for n=2
    <br />
P((E_1 \cup E_2 \cup ... \cup E_n) \cup E_{n+1}) = 1-P((E_1^{C} \cap E_2^{C} \cap ... \cap E_n^{C}) \cap E_{n+1}) = 1- P(E_1^{C} \cap E_2^{C} \cap ... \cap E_n^{C})P(E_{n+1}^{C})

    But by induction hypothesis
    1- P(E_1^{C} \cap E_2^{C} \cap ... \cap E_n^{C}) = 1-P(E_1^{C})*P(E_2^{C})*...*P(E_n^{C})

    Thus
    P(E_1 \cup E_2 \cup ... \cup E_n \cup E_{n+1}) = P((E_1 \cup E_2 \cup ... \cup E_n) \cup E_{n+1}) = 1-P(E_1^{C})*P(E_2^{C})*...*P(E_n^{C})*P(E_{n+1}^C})

    Which is what we want.

    I was wondering if I proved the base case correctly and applied the induction hypothesis correctly. Thank you for your help.
    Last edited by Jame; March 10th 2011 at 07:15 AM. Reason: laty6ex editing
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