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

  1. #1
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    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 ^
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by CaptainLorax View Post
    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:

    E(X|X>2)=E(X)+2

    and:

    \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
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  3. #3
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    thank you very much
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  4. #4
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    actually I still don't understand how the end of your equation = 1/(1/6)
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by CaptainLorax View Post
    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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  6. #6
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    now I just have to figure out how to send a virtual thanks your way
    edit-found it
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