# Poisson

• March 17th 2011, 03:26 PM
dwsmith
Poisson
$\lambda=4$

$\displaystyle P(X\geq 5|X\geq 2)=\frac{P(X\geq 5\cap X\geq 2)}{P(X\geq 2)}=\frac{P(X\geq 5)}{P(X\geq 2)}=\frac{1-P(X<5)}{1-P(X<2)}$

When I did this, the answer was incorrect. Is this process wrong and why?
• March 17th 2011, 03:43 PM
pickslides

I get $\displaystyle \frac{1-\left(\frac{4^4e^{-4}}{4!}+\frac{4^3e^{-4}}{3!}+\dots +\frac{4^0e^{-4}}{0!}\right)}{1-\left(\frac{4^1e^{-4}}{1!} +\frac{4^0e^{-4}}{0!}}\right)$
• March 17th 2011, 03:45 PM
dwsmith
Quote:

Originally Posted by pickslides

I get $\displaystyle \frac{1-\left(\frac{4^4}{4!}+\frac{4^3}{3!}+\dots +\frac{4^0}{0!}\right)}{1-\left(\frac{4^2}{2!}+\frac{4^1}{1!} +\frac{4^0}{0!}}\right)$

You forgot $e^{-4}$, but I obtain .4085 and the answer is .37 somethingish.
• March 17th 2011, 03:51 PM
pickslides
Quote:

Originally Posted by dwsmith
You forgot $e^{-4}$,

I did forget, but stick by my answer (after the edit!). Sorry dw, might have to wait for another forum member to come through and eyeball the problem.

Maybe the book's answer is wrong?
• March 17th 2011, 04:58 PM
awkward
Quote:

Originally Posted by dwsmith
$\lambda=4$

$\displaystyle P(X\geq 5|X\geq 2)=\frac{P(X\geq 5\cap X\geq 2)}{P(X\geq 2)}=\frac{P(X\geq 5)}{P(X\geq 2)}=\frac{1-P(X<5)}{1-P(X<2)}$

When I did this, the answer was incorrect. Is this process wrong and why?

Looks right to me.

I get 0.4086 as the answer.