Mixing
Uniform and exponential distribution
$begingroup$
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
asked Jan 14 at 13:03
pawelKpawelK
527
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add a comment |
$begingroup$
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
asked Jan 14 at 13:03
pawelKpawelK
527
$endgroup$
add a comment |
$begingroup$
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
asked Jan 14 at 13:03
pawelKpawelK
527
$endgroup$
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
asked Jan 14 at 13:03
pawelKpawelK
527
asked Jan 14 at 13:03
pawelKpawelK
527
asked Jan 14 at 13:03
pawelKpawelK
527
asked Jan 14 at 13:03
pawelKpawelK
527
asked Jan 14 at 13:03
pawelKpawelK
527
527
add a comment |
add a comment |
1 Answer
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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*}
answered Jan 14 at 13:20
P. QuintonP. Quinton
1,8011213
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1 Answer
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1 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*}
answered Jan 14 at 13:20
P. QuintonP. Quinton
1,8011213
$endgroup$
add a comment |
$begingroup$
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*}
answered Jan 14 at 13:20
P. QuintonP. Quinton
1,8011213
$endgroup$
add a comment |
$begingroup$
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*}
answered Jan 14 at 13:20
P. QuintonP. Quinton
1,8011213
$endgroup$
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*}
answered Jan 14 at 13:20
P. QuintonP. Quinton
1,8011213
answered Jan 14 at 13:20
P. QuintonP. Quinton
1,8011213
answered Jan 14 at 13:20
P. QuintonP. Quinton
1,8011213
answered Jan 14 at 13:20
P. QuintonP. Quinton
1,8011213
1,8011213
add a comment |
add a comment |
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