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Thread: The office

  1. #1
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    Question The office

    Office gets each day new mail according to Poisson distribution $\displaystyle P(\lambda)$.
    Days are independent.
    Each day, one mail gets answered (if there is any), other mails (if there are any) are left for later days.
    What is the probability, that after 2 days, the office is left with at least 2 un-answered mails?
    So, we're looking for $\displaystyle P(X\geq 2)$ ...

    $\displaystyle P(X\geq 2)=1-P(X=0)-P(X=1)=1-...$

    P(X=0) - 5 ways to be left with 0 mail at day 3 (# income mail on day 1, # income mail on day 2): (0,0) (0,1) (1,0) (1,1) (2,0)

    P(X=1) - 3 ways to be left with 1 mail at day 3: (0,2) (1,2) (2,1)

    Now, how do you use above with Poisson formula?

    Official solution is $\displaystyle 1-e^{-2\lambda}(1+2\lambda+2\lambda^2+\frac{7\lambda^3}{ 6})$.
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  2. #2
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    according to wikipedia, the pmf for a poisson variable W is $\displaystyle P(W=w) = \frac{ \lambda^k}{k!}e^{-\lambda} $

    You are looking at 2 independant poisson variables (Y,Z). Just multiply P(Y=y) * P(Z=z)
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  3. #3
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    I am really unable to make this one on my own.
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  4. #4
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    for exasmple, the (0,1) scenario:

    Suppose Y is the number of mails on the first day, and Z is the number of mails on the second day

    P(0,1)=P(Y=0 and Z=1)

    $\displaystyle \displaystyle =P(Y=0) \times P(Z=1) = \left( e^{-\lambda} \times \frac{\lambda^0}{0!} \right) \times \left( e^{-\lambda} \times \frac{\lambda^1}{1!} \right) $
    $\displaystyle = \displaystyle \left( e^{-\lambda} \times \frac{1}{1} \right) \times \left( e^{-\lambda} \times \frac{\lambda}{1} \right)$

    $\displaystyle = \displaystyle e^{-\lambda} (1+\lambda)$



    use the same approach for the other scenarios you identified
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  5. #5
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    Got it ... but your last step should be $\displaystyle e^{-2\lambda}\lambda$ anyways.
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