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Thread: Probability System, discrete dist.

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    Probability System, discrete dist.




    here is the system we're dealing with here


    Component 1 is the key one while an identical backup component is 2 and is in an "off" state until needed.
    The length of life of the component Xi, i=1,2

    Okay so the discrete distribution is given..
    \begin{matrix}
x &     0  & 1  & 2 \\
p(x) &  .1 & .5 & .4 \\
\end{matrix}

    I am attempting to find the cdf and pdf of a lifetime X of a system.
    I thought i would treat this as a normal system but the fact that it is not parallel nor in series but instead is a Standby model is confusing me.
    If i treat this as a normal system I got

    \begin{matrix}
X      & 2 &           3 &        4 &    5 &    6 \\
p(X) & (0.1)^2 & 2(.1)(.5) \\
\end{matrix}

    just need some guidance with this....

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    Re: Probability System, discrete dist.

    I guess a few things just aren't clear to me.

    You say "The length of life of the component Xi, i=1,2" and that's the end of your sentence. Do you mean that the following table represents the probabilities of the lifetimes? That's what I assume but it wasn't clear.

    Also, If the individuals can have a lifetime of 0 why is the minimum lifetime of the system 2? It just doesn't seem consistent to me. If you could explain that then maybe I'll help you out a little bit.

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    Re: Probability System, discrete dist.

    In series: minimum of the lifetime
    In parallel: maximum of the lifetime
    standby mode: sum of the lifetime

    So I wonder the support should be \{0, 1, 2, 3, 4\} instead,
    as Dason suggested.

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    Re: Probability System, discrete dist.

    The length of life of a component Xi, i=1,2, is measured in thousands of hours and has the discrete distribution that i posted above.

    thats all the problem gives me, but i dont understand why x={0,1,3} and not just {1,2}


    when you say in series is the mininum of the lifetime, does that mean 1 and 2 are in series or 1 is just in series by itself? The fact that the problem says 2 is a backup component is messing me up

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    Re: Probability System, discrete dist.

    I believe that just means that the system won't activate 2 until 1 has broke. Think about buying two new light bulbs. They don't go bad just sitting there (theoretically). Let's say you only have one lamp in your house. So you put lightbulb1 in and it's sitting there all happy doing it's job and then one day it breaks. So you throw out lightbulb1 and you replace it with lightbulb2. That's essentially the system you have going on here.

    A system that has components in series: Pretend you want to go to your friends house and there is only one road that gets there. Along this road there are two bridges. If bridge 1 breaks then you can't get to your friends house. If bridge two breaks then you can't get to your friends house. Both bridges need to be in tact to get there. The route to your friend's house depends on the series of bridges and the state that they're in.

    Hopefully this helps you understand BGM's post a little more.

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    Re: Probability System, discrete dist.

    okay that makes more sense, but if it a discrete distribution, how are you suppose to find the cdf or pdf?

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    Re: Probability System, discrete dist.

    Well it's discrete so you can just examine all the possible outcomes.

    X1 = 0, X2 = 0
    X1 = 0, X2 = 1
    X1 = 0, X2 = 2
    X1 = 1, X2 = 0
    ...

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    Re: Probability System, discrete dist.

    so would need to find the pdf for X and the pdf for Y seperately or should i find the joint pdf?

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    Re: Probability System, discrete dist.

    Now you are interested in the probability mass function of the sum of the two lifetime
    random variables.

    So let X = X_1 + X_2

    You can fully list out the all possible cases (3 * 3 = 9 cases total)
    and then count in which case corresponding to the sum = 0, 1, 2, 3, 4
    And you assume they are independent, so you can easily find out the corresponding
    joint probability.

    I do 1 example for you:
    \Pr\{X = 0\} = \Pr\{X_1 + X_2 = 0\} = \Pr\{X_1 = 0\}\Pr\{X_2 = 0\}


    PS:
    Note for discrete random variable,
    it should be probability mass function (pmf) instead of probability density function (pdf)

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    Re: Probability System, discrete dist.

    so P(X=0)=(.1)(.1) i used the distribution table..

    then P(X=1)=P(x1=0)P(x2=1)=(.1)(.5)

    am i on the right trail?

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    Re: Probability System, discrete dist.

    Quote Originally Posted by gator78 View Post
    so P(X=0)=(.1)(.1) i used the distribution table..

    then P(X=1)=P(x1=0)P(x2=1)=(.1)(.5)

    am i on the right trail?
    Well... close?

    P(X = 1) =
    P( X1 + X2 = 1) =
    P( {X1 = 1, X2 = 0} or {X1 = 0, X2 = 1}) =
    P(X1 = 1, X2 = 0) + P(X1 = 0, X2 = 1) =
    P(X1 = 1)P(X2 = 0) + P(X1 = 0)P(X2 = 1) =
    (.5)(.1) + (.1)(.5) =
    2(.1)(.5) =
    .1

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    Re: Probability System, discrete dist.

    i set up a table with X1 and X2 making the marginal probabilities the discrete dist. i was given. then where (X1,X2) is (0,1) or (1,0) i find the probabilites and sum them...
    i think

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    Re: Probability System, discrete dist.

    You should be able to form a table that recording the sum
    X_1 + X_2 like this:

    \begin{tabular}{|c||c|c|c|} \hline
& 0 & 1 & 2 \\ \hline \hline
0 & 0 & 1 & 2 \\ \hline
1 & 1 & 2 & 3  \\ \hline
2 & 2 & 3 & 4 \\ \hline
\end{tabular}

    That will ensure you will not miscount the cases.
    And then just add up the joint probabilities of the corresponding entries.

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    Re: Probability System, discrete dist.

    i found the pmf for the discrete distribution
    \begin{matrix}
X     &  0 &  1&   2&   3&    4&   6 \\
p(x) &  .01 &.1& .25&.08 &.4   &.16\\
\end{matrix}
    5 is not in there because of the distribution given 5 is impossible..

    and found the EX=3.4 (just multiply the X and p(x) and sum for all of them)

    I know this may be simple but i am now trying to find the cdf....
    I am pretty sure i dont need a table but more like functions for given values of X correct.
    for example the cdf so far that I have is.....
    0.01 if 0<=x<=1 and 0.1 if 1<=x<=2... etc.

    am i doing this right for the cdf? i also thought about drawing it but im not sure..

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    Re: Probability System, discrete dist.


    Wait... How is 5 not possible but 6 is? Also your cdf doesn't look right even for the distribution you gave. It looks like you're giving a wrong version of the pdf. Remember the definition of the cdf. F(x) = P(X <= x). So in this case F(1) = P(X=0) + P(X=1), F(2) = P(X=0) + P(X=1) + P(X=2), ...
    Last edited by Dason; 11-10-2010 at 10:56 PM.

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