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

Aug 2010
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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 ^
 

CaptainBlack

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Nov 2005
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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
 
Aug 2010
4
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actually I still don't understand how the end of your equation = 1/(1/6)
 

CaptainBlack

MHF Hall of Fame
Nov 2005
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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
 
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Aug 2010
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now I just have to figure out how to send a virtual thanks your way
edit-found it