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Proving complement of unions equals intersection of complements.

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Generalize to obtain (C1 U C2 U...U Ck)' = C1' intersect C2' intersect...intersect Ck'

    ' = complement

    Say that C1, C2,...,Ck are independent events that have respective probabilities p1, p2, ..., pk. Argue that the probability of at least one of C1, C2,...,Ck is equal to 1 - (1-p1)(1-p2)...(1-pk)

    2. Relevant equations

    I don't know how to generalize that...

    For the second part, P(C1 U C2 U...U Ck) = 1- P(C1 U C2 U...U Ck)' = 1 - P(C1' intersect C2' intersect...intersect Ck') = 1 - (1-p1)(1-p2)...(1-pk). Not sure how that proves at least one of Ck has to equal that though...

    3. The attempt at a solution
     
  2. jcsd
  3. Sep 30, 2009 #2

    statdad

    User Avatar
    Homework Helper

    for the first part why not begin trying to show it is true for k = 2: that is, try to show

    [tex]
    (C_1 \cup C_2)' = C_1' \cap C_2'
    [/tex]

    Once you have it for k=2, use induction for the general case.

    for the second part (once the first is shown) your first line should read

    [tex]
    \Pr(C_1 \cup C_2 \cup \cdots \cup C_k) = 1 - \Pr((C_1 \cup C_2 \cup \cdots C_k)') = 1 - \Pr(C_1' \cap C_2' \cap \cdots \cap C_k')
    [/tex]

    At this point, use the facts that [tex] \Pr(C_j) = p_j [/tex] (so you know the probabilities of the complements) as well as the fact that the events are independent.
     
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