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Math Help - Probability pdf

  1. #1
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    Probability pdf

    Hey all. Here's another I can't get.

    The life, X in years of a car part had pdf:

    f(x)= 3x / 7 * e^-(x/7) ; 0 < x < infinity

    a) probability that this regulator will last at least 7 years?
    b) Given that it lasts 7 yrs, conditional prob. that it will last at least another 3.5 yrs?


    ------

    So for a, I got the correct answer by saying p(x>=7) = 1-p(x<=7)
    And integrating f(x) from 0--->7

    The answer is 0.368

    Now for b, would I say 1-integral from 7--> 10.5? Cuz' that's what I was thinking, but it didn't work out. Please explain!
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  2. #2
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    So you want  P(X \geq 10.5| X = 7) . What is the definition of conditional probability?


    Quote Originally Posted by amor_vincit_omnia View Post
    Hey all. Here's another I can't get.

    The life, X in years of a car part had pdf:

    f(x)= 3x / 7 * e^-(x/7) ; 0 < x < infinity

    a) probability that this regulator will last at least 7 years?
    b) Given that it lasts 7 yrs, conditional prob. that it will last at least another 3.5 yrs?


    ------

    So for a, I got the correct answer by saying p(x>=7) = 1-p(x<=7)
    And integrating f(x) from 0--->7

    The answer is 0.368

    Now for b, would I say 1-integral from 7--> 10.5? Cuz' that's what I was thinking, but it didn't work out. Please explain!
    Last edited by particlejohn; July 4th 2008 at 09:39 PM.
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  3. #3
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    The intersection of the two/P(x>=10.5)


    How do I get the intersection, the product of the two?
    And for p(x>=10.5), do I do 1-p(x<=10.5) from 0-->10.5 ?
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  4. #4
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    actually,  P(X \geq 10.5| X = 7) = \frac{P( X \geq 10.5 \ \text{and} \ X = 7)}{P(X = 7)} . So  X \geq 10.5 and  X = 7 are mutually exclusive events. So I think it would be  \frac{P(X \geq 10.5)}{P(X=7)} . You can't take the product, because they are not independent.


    Quote Originally Posted by amor_vincit_omnia View Post
    The intersection of the two/P(x>=10.5)


    How do I get the intersection, the product of the two?
    And for p(x>=10.5), do I do 1-p(x<=10.5) from 0-->10.5 ?
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  5. #5
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    Thanks, it worked..
    However, I thought that if two events were mutually exclusive, their intersection is 0?

    So in that formula, how can you just say to divide the p(x>=10.5/x=7) ?
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  6. #6
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    given the context,  X is the lifetime of a car. So if  X \geq 10.5 then it must have reached  X = 7 . So they are not really mutually exclusive. They couldn't be, because then the conditional probability would not make sense.

    what was the answer given? was it correct?
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  7. #7
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    yea, it was. I thanked" you..lol

    I appreciate it
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