Probability of x failing before y

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To find the probability that component X fails before component Y, calculate P{X<Y} using the joint density functions f(x) = exp(-x) and f(y) = 2*exp(-2y). This involves integrating over the region A = {(x,y) | 0 ≤ x ≤ y}. The probability can be expressed as the double integral ∫∫_A 2e^(-x)e^(-2y) dx dy, which can be separated due to the independence of X and Y. The next step is determining the appropriate limits for the integral to complete the calculation.
tony3333
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hi,
i have a problem and i really want you to help me with it.
if we have: density of X : f(x)=exp(-x), and density of Y : f(y)=2*exp(-2y) (independent components), what is the probability that X component fails first?
(it should be a number)
thank you
 
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If I understand correctly, you will want P{X<Y}.

You can calculate this as follows: let A=\{(x,y)~\vert~0\leq x\leq y\}
Then you need to calculate

\int\int_A{2e^{-x}e^{-2y}dxdy}

this will be your probability...
 
micromass said:
If I understand correctly, you will want P{X<Y}.

You can calculate this as follows: let A=\{(x,y)~\vert~0\leq x\leq y\}
Then you need to calculate

\int\int_A{2e^{-x}e^{-2y}dxdy}

this will be your probability...

and because these are independent, we can separate the integrals and have

\int_A{e^{-x}dx \int_A{2e^{-2y}dy

and then the variables of every probability , x,y, can be descibed by the same symbol, let's say t
but what should be the limits of the integral?
 
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I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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