On my studying material I noticed that for "devices in parallel" problems, a question of the type "Find the probability the device works for 1 year" is dealt with P(A or B) where P(A), P(B) are the probabilities of each of the instruments working for one year, which is something I intuitively understand.(adsbygoogle = window.adsbygoogle || []).push({});

Now, when I went to a random variables chapter problem, for two random variables X and X, given their probability density function, for two devices in parallel where each device's life is represented by each random variable, the question "find the probability the device works for between 1 and 2 years" is dealt with the probability of max{X, Y} satisfying that criteria which appears to give a different result than the union of the corresponding probabilities!

I'll try to translate the actual problem: "The lifetimes X,Y (in hours) of two components are random variables with common probability density function fx,y(x,y) = e^(-x-y) with x,y>0. Calculate the probability the lifetime of the machine is between 1 and 2 hours when the components are connected in parallel." (it also asks for a connection in series). (It is also given from a previous exercise that the random variables are independent.)

My understanding is that if I use fx and fy and find the components probabilities separately and then calculate their union, the result would be the same. But in this case the result is: P(1<X<2)=P(1<Y<2)=0.2325 with P(1<X<2 or 1<Y<2) = 0.4109

While if I actually find the function for max{X,Y}, it's P(max{X,Y}<z) = P(X<z,Y<z) = P(X<z)P(Y<z) = Fx(x)Fy(y) which gives a different result.

Are the approaches equivalent?

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# Are those solutions equivalent?

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