$\displaystyle \lambda=4$

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

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

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- Mar 17th 2011, 03:26 PMdwsmithPoisson
$\displaystyle \lambda=4$

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

When I did this, the answer was incorrect. Is this process wrong and why? - Mar 17th 2011, 03:43 PMpickslides
Looks correct to me. Was your answer even close?

I get $\displaystyle \displaystyle \frac{1-\left(\frac{4^4e^{-4}}{4!}+\frac{4^3e^{-4}}{3!}+\dots +\frac{4^0e^{-4}}{0!}\right)}{1-\left(\frac{4^1e^{-4}}{1!} +\frac{4^0e^{-4}}{0!}}\right)$ - Mar 17th 2011, 03:45 PMdwsmith
- Mar 17th 2011, 03:51 PMpickslides
- Mar 17th 2011, 04:58 PMawkward