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Math Help - The office

  1. #1
    Member courteous's Avatar
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    Question The office

    Office gets each day new mail according to Poisson distribution 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 P(X\geq 2) ...

    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 1-e^{-2\lambda}(1+2\lambda+2\lambda^2+\frac{7\lambda^3}{  6}).
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  2. #2
    MHF Contributor
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    according to wikipedia, the pmf for a poisson variable W is 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
    Member courteous's Avatar
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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 =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  \left( e^{-\lambda} \times \frac{1}{1} \right) \times \left( e^{-\lambda} \times \frac{\lambda}{1} \right)

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



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