# Thread: X = # of times u need to roll to get the first 5. E(X ∣ X > 2) = (1/(1/6)) + 2 Why?

1. ## X = # 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 ^

2. Originally Posted by CaptainLorax
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 ^
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

3. thank you very much

4. actually I still don't understand how the end of your equation = 1/(1/6)

5. Originally Posted by CaptainLorax
actually I still don't understand how the end of your equation = 1/(1/6)
To get the first 5 on the n-th roll requires that you roll n-1 non-fives with probability 5/6 and the last roll is a 5 with probability 1/6

CB

6. now I just have to figure out how to send a virtual thanks your way
edit-found it