Hello, this is my first time posting here. I am studying for the P1 actuarial exam

Thank you for any help :)

X = # of times you need to roll to get the first 5.

E(X ∣ X > 2) = (1/(1/6)) + 2

How do I get this ^ solution ^

- Aug 22nd 2010, 05:46 PMCaptainLoraxX = # of times u need to roll to get the first 5. E(X ∣ X > 2) = (1/(1/6)) + 2 Why?
Hello, this is my first time posting here. I am studying for the P1 actuarial exam

Thank you for any help :)

X = # of times you need to roll to get the first 5.

E(X ∣ X > 2) = (1/(1/6)) + 2

How do I get this ^ solution ^ - Aug 22nd 2010, 07:31 PMCaptainBlack
Lets assume you are talking rolling a fair die. The prior history of rolls does not effect the future outcomes, so:

$\displaystyle E(X|X>2)=E(X)+2$

and:

$\displaystyle \displaystyle E(X)=\sum_{k=1}^{\infty} k \left(\frac{5}{6}\right)^{k-1}\frac{1}{6}=\frac{1}{6} \sum_{k=1}^{\infty} k \left(\frac{5}{6}\right)^{k-1}$

CB - Aug 22nd 2010, 08:36 PMCaptainLorax
thank you very much

- Aug 22nd 2010, 10:10 PMCaptainLorax
actually I still don't understand how the end of your equation = 1/(1/6)

- Aug 24th 2010, 02:19 AMCaptainBlack
- Aug 24th 2010, 11:16 AMCaptainLorax
now I just have to figure out how to send a virtual thanks your way

edit-found it