1. ## Poisson

$\displaystyle \lambda=4$

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

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

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

3. Originally Posted by pickslides

I get $\displaystyle \displaystyle \frac{1-\left(\frac{4^4}{4!}+\frac{4^3}{3!}+\dots +\frac{4^0}{0!}\right)}{1-\left(\frac{4^2}{2!}+\frac{4^1}{1!} +\frac{4^0}{0!}}\right)$
You forgot $\displaystyle e^{-4}$, but I obtain .4085 and the answer is .37 somethingish.

4. Originally Posted by dwsmith
You forgot $\displaystyle e^{-4}$,
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?

5. Originally Posted by dwsmith
$\displaystyle \lambda=4$

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

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.