The office

**courteous** · August 27th 2010, 09:18 AM

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 $1-e^{-2\lambda}(1+2\lambda+2\lambda^2+\frac{7\lambda^3}{ 6})$ .

**SpringFan25** · August 27th 2010, 09:51 AM

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)

**courteous** · August 27th 2010, 12:08 PM

I am really unable to make this one on my own.

**SpringFan25** · August 27th 2010, 02:04 PM

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

**courteous** · August 28th 2010, 01:34 AM

Got it ... but your last step should be $e^{-2\lambda}\lambda$ anyways.

Thread: The office

LinkBack

Thread Tools

Display

The office

Similar Math Help Forum Discussions

Latex plug in for microsoft office

mathtype help with office '07

LaTeX for MS office?

Office version along with Window Vista