Uniform and exponential distribution












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Consider an experiment. The duration of the experiment has uniform distribution on $[2,6]$h. When the experiment starts, the device A turns on. This device will turn off after time,which has exponential distribution with $lambda=1/4$. Calculate the probability, that experiment will end, before the device A is turned off.



Let:
$Y$- when the device A turns off.
Do we have to calculate $P(0 le Y le 6)$ or $P(2 le Y le 6)$?
My idea is to count the integral :
$int_{?}^{6} int_{0}^{frac{e^{-x/4}}{4}}1 ,dx,dy$.






probability-theory uniform-distribution exponential-distribution







share|cite|improve this question









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    1












    $begingroup$


    Consider an experiment. The duration of the experiment has uniform distribution on $[2,6]$h. When the experiment starts, the device A turns on. This device will turn off after time,which has exponential distribution with $lambda=1/4$. Calculate the probability, that experiment will end, before the device A is turned off.



    Let:
    $Y$- when the device A turns off.
    Do we have to calculate $P(0 le Y le 6)$ or $P(2 le Y le 6)$?
    My idea is to count the integral :
    $int_{?}^{6} int_{0}^{frac{e^{-x/4}}{4}}1 ,dx,dy$.






    probability-theory uniform-distribution exponential-distribution







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Consider an experiment. The duration of the experiment has uniform distribution on $[2,6]$h. When the experiment starts, the device A turns on. This device will turn off after time,which has exponential distribution with $lambda=1/4$. Calculate the probability, that experiment will end, before the device A is turned off.



      Let:
      $Y$- when the device A turns off.
      Do we have to calculate $P(0 le Y le 6)$ or $P(2 le Y le 6)$?
      My idea is to count the integral :
      $int_{?}^{6} int_{0}^{frac{e^{-x/4}}{4}}1 ,dx,dy$.






      probability-theory uniform-distribution exponential-distribution







      share|cite|improve this question









      $endgroup$




      Consider an experiment. The duration of the experiment has uniform distribution on $[2,6]$h. When the experiment starts, the device A turns on. This device will turn off after time,which has exponential distribution with $lambda=1/4$. Calculate the probability, that experiment will end, before the device A is turned off.



      Let:
      $Y$- when the device A turns off.
      Do we have to calculate $P(0 le Y le 6)$ or $P(2 le Y le 6)$?
      My idea is to count the integral :
      $int_{?}^{6} int_{0}^{frac{e^{-x/4}}{4}}1 ,dx,dy$.






      probability-theory uniform-distribution exponential-distribution




      probability-theory uniform-distribution exponential-distribution






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 14 at 13:03









      pawelKpawelK

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          $begingroup$

          I suppose that the exponential distribution is independent of the end of the experiment. Let $X$ be the end of the experiment and $Y$ be the time at which the device turns of. You are looking at the probability of the event $Xleq Y$, that is
          begin{align*}
          mathbb P[Xleq Y] &= int_2^6 int_x^infty f_{X,Y}(x,y) dy dx\
          &= int_2^6 int_x^infty frac{1}{4}cdot frac{1}{4}e^{-frac{1}{4}y} dy dx\
          &= frac{1}{4}int_2^6 e^{-frac{1}{4} x} dx\
          &= e^{-frac{1}{2}} - e^{-frac{3}{2}}
          end{align*}






          share|cite|improve this answer









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            $begingroup$

            I suppose that the exponential distribution is independent of the end of the experiment. Let $X$ be the end of the experiment and $Y$ be the time at which the device turns of. You are looking at the probability of the event $Xleq Y$, that is
            begin{align*}
            mathbb P[Xleq Y] &= int_2^6 int_x^infty f_{X,Y}(x,y) dy dx\
            &= int_2^6 int_x^infty frac{1}{4}cdot frac{1}{4}e^{-frac{1}{4}y} dy dx\
            &= frac{1}{4}int_2^6 e^{-frac{1}{4} x} dx\
            &= e^{-frac{1}{2}} - e^{-frac{3}{2}}
            end{align*}






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              I suppose that the exponential distribution is independent of the end of the experiment. Let $X$ be the end of the experiment and $Y$ be the time at which the device turns of. You are looking at the probability of the event $Xleq Y$, that is
              begin{align*}
              mathbb P[Xleq Y] &= int_2^6 int_x^infty f_{X,Y}(x,y) dy dx\
              &= int_2^6 int_x^infty frac{1}{4}cdot frac{1}{4}e^{-frac{1}{4}y} dy dx\
              &= frac{1}{4}int_2^6 e^{-frac{1}{4} x} dx\
              &= e^{-frac{1}{2}} - e^{-frac{3}{2}}
              end{align*}






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                I suppose that the exponential distribution is independent of the end of the experiment. Let $X$ be the end of the experiment and $Y$ be the time at which the device turns of. You are looking at the probability of the event $Xleq Y$, that is
                begin{align*}
                mathbb P[Xleq Y] &= int_2^6 int_x^infty f_{X,Y}(x,y) dy dx\
                &= int_2^6 int_x^infty frac{1}{4}cdot frac{1}{4}e^{-frac{1}{4}y} dy dx\
                &= frac{1}{4}int_2^6 e^{-frac{1}{4} x} dx\
                &= e^{-frac{1}{2}} - e^{-frac{3}{2}}
                end{align*}






                share|cite|improve this answer









                $endgroup$



                I suppose that the exponential distribution is independent of the end of the experiment. Let $X$ be the end of the experiment and $Y$ be the time at which the device turns of. You are looking at the probability of the event $Xleq Y$, that is
                begin{align*}
                mathbb P[Xleq Y] &= int_2^6 int_x^infty f_{X,Y}(x,y) dy dx\
                &= int_2^6 int_x^infty frac{1}{4}cdot frac{1}{4}e^{-frac{1}{4}y} dy dx\
                &= frac{1}{4}int_2^6 e^{-frac{1}{4} x} dx\
                &= e^{-frac{1}{2}} - e^{-frac{3}{2}}
                end{align*}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 13:20









                P. QuintonP. Quinton

                1,8011213




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