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Probabilities and independence

  1. Oct 16, 2004 #1
    Hi...Can you plz check if my proof is correct?

    A1,A2,.....An are independently events.
    Prove that :
    P(A1[union]A2[union]...[union]An) = 1-Πi[element-of]I(1-P(Ai))

    note for this (Πi[element-of]I(1-P(Ai))
    P([intersect]Ai)= Π P(Ai)
    for 3 events A1,A2,A3
    means: P(A1[intersect]A2)=P(A1)*P(A2)
    P(A1[intersect]A2[intersect]A3)=P(A1)* P(A2) * P(A3)

    Now my proof:
    We know that P([intersect]Ai)= Π P(Ai)
    if A1,A2,...,An are independent then and the complements
    are independent

    P([intersect]Ai)complement = Π P(Aicomplement)
    P([union](Ai compl) ) = Π(1-P(Ai))
    1-P([union]Ai)= Π(1-P(Ai))
    Finally ... we got our proof
    Is it correct?

    And one more....
    but i dont know how to prove this:
    A,B,C are independent
    We must prove that A and B[union]C are independent too
  2. jcsd
  3. Oct 17, 2004 #2


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    I couldn't follow some steps in your derivation (then again, it's pretty early here and I`m half asleep).

    (Try using LateX, it makes your notation look more sexy. Click on the formulas to see the code.)

    I would do the following:
    [tex]P\left(\bigcup_{i=1}^n A_i\right)=1-P\left((\bigcup_{i=1}^n A_i)^c\right)=1-P\left(\bigcap_{i=1}^n A_i^c\right)=1-\prod_{i=1}^nP\left(A_i^c\right)=1-\prod_{i=1}^n(1-P\left(A_i)\right)[/tex]
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