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Math Help - Probability That None of Independent Events Occurs

  1. #1
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    Probability That None of Independent Events Occurs

    The problem says:

    Let A_1, A_2, A_3,...,A_n be independent events. Show that the probability that none of the A_i's occur is less than or equal to exp(- \Sigma_{i=1}^{n}P(A_i)).

    I'm confused at where the inequality and the exponent are coming from.

    Here is what i've done.

    I figured that the P(none happening) = 1 - P(one happening)

    P(one happening) = \Sigma_{k=1}^{n}P(A_k).

    Then, P(none) = 1 - \Sigma_{k=1}^{n}P(A_k).

    I think this is right, but where to go from here, I'm not sure.

    Thank you!
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  2. #2
    MHF Contributor matheagle's Avatar
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    First of all
    P(none happening) = 1 - P(at least one happening)

    Then P(at least one happening)

    =P(\cup A_i)\le\sum_{i=1}P(A_i)

    BUT by independence this is an equality (which is why it works).

    What you seem to need is that 1-x\le e^{-x}

    Use calculus to prove that.

    Let f(x)=e^{-x}+x-1

    then obtain the first two derivatives and show that the min is 0 at 0, so f(x)\ge 0.
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  3. #3
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    So, the only thing I had wrong was that it should be P(at least one happening) rather than just P(one happening), but that should still be equal to , correct?

    I will try to work out the other stuff today. Thank you!
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  4. #4
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    That is correct.

    Now all you need to show is that

     1- \sum_{k=1}^{n}P(A_k) \leq e^{-\sum_{k=1}^{n}P(A_k)}

    which is equivalent to proving, as matheagle explained:

     1- x \leq e^{-x}

    EDIT: Thanks matheagle, I'm always messing up =X
    Last edited by pedrosorio; September 2nd 2009 at 10:59 AM.
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  5. #5
    MHF Contributor matheagle's Avatar
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    you're missing the minus in the exponent
    but that's true too

     1- x \leq e^x

    we want  1- x \leq e^{-x}

    Quote Originally Posted by pedrosorio View Post
    That is correct.

    Now all you need to show is that

     1- \sum_{k=1}^{n}P(A_k) \leq e^{\sum_{k=1}^{n}P(A_k)}

    which is equivalent to proving, as matheagle explained:

     1- x \leq e^x
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  6. #6
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    Thank you both very much! I got it now
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