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Thread: General Probability Problem (Actuary question)

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    General Probability Problem (Actuary question)

    https://imgur.com/a/nMeLwip



    Not sure what to do with this problem:

    I know that the difference is the same for p_(0)-p_(1)=p_(1)-p_(2)=p_(2)-p_(3)=p_(3)-p_(4)=p_(4)-p_(5)


    I know we are looking for p_(4)+p_(5)
    Last edited by math951; Oct 5th 2018 at 03:35 PM.
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    Re: General Probability Problem (Actuary question)

    Quote Originally Posted by math951 View Post
    https://imgur.com/a/nMeLwip
    Not sure what to do with this problem:
    I know that the difference is the same for p_(0)-p_(1)=p_(1)-p_(2)=p_(2)-p_(3)=p_(3)-p_(4)=p_(4)-p_(5)
    I know we are looking for p_(4)+p_(5)
    Do you know that $\displaystyle \sum\limits_{k = 0}^5 {{p_k}} = 1~?$ If so explain.

    If $\delta=p_0-p_1$ is it true that $p_n=\delta+p_{n+1},~\;n=1,2,3,4~?$ Can you find $\delta~?$
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    Re: General Probability Problem (Actuary question)

    I would approach it as an arithmetic progression where a is P(0) and the common difference is d.

    So the terms are a, a+d, a+2d, a+3d, a+4d, a+5d.


    All of these add to 1 (total of probabilities). The first 2 add to 0.4. This will give 2 simultaneous equations. Then find sum of last 2.
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    Re: General Probability Problem (Actuary question)

    Quote Originally Posted by Plato View Post
    Do you know that $\displaystyle \sum\limits_{k = 0}^5 {{p_k}} = 1~?$ If so explain.

    If $\delta=p_0-p_1$ is it true that $p_n=\delta+p_{n+1},~\;n=1,2,3,4~?$ Can you find $\delta~?$
    Doesn't the probability space always have to equal one?
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    Re: General Probability Problem (Actuary question)

    Quote Originally Posted by Debsta View Post
    I would approach it as an arithmetic progression where a is P(0) and the common difference is d.

    So the terms are a, a+d, a+2d, a+3d, a+4d, a+5d.


    All of these add to 1 (total of probabilities). The first 2 add to 0.4. This will give 2 simultaneous equations. Then find sum of last 2.
    You have the progression backwards.

    $$p_0=p_5+5d, p_1=p_5+4d, p_2=p_5+3d, p_3=p_5+2d, p_4=p_5+d, p_5=p_5$$
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    Re: General Probability Problem (Actuary question)

    Quote Originally Posted by SlipEternal View Post
    You have the progression backwards.

    $$p_0=p_5+5d, p_1=p_5+4d, p_2=p_5+3d, p_3=p_5+2d, p_4=p_5+d, p_5=p_5$$
    Not really, it's just that my d will be negative.
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    Re: General Probability Problem (Actuary question)

    Quote Originally Posted by Debsta View Post
    Not really, it's just that my d will be negative.
    Oh, true. My bad.
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