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I have a system with components: A and B. The operating times until failure of two are independent and exponentially distributed with $A \sim \mathrm{Exp}(2)$ and $B \sim \mathrm{Exp}(3)$. Assumption given: the system fails at the first component failure. Suppose that component $A$ fails first. Find the mean remaining operating life of component B.

So my attempt is to use the tail probability formula to calculate the expectation of this.

I first find the probability that component $A$ fails first which is $P(X_{A} < X_{B}) = 2/5 $ by using the joint probability mass function:

$ f_{A,B}(a,b) = 6 e^{-2a} e^{-3b} \quad a,b > 0$.

Is there another way I could approach this for the mean remaining life of component B given component A fails which is $(X_{A} < X_{B})$?

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  • $\begingroup$ By the memoryless property of the exponential distribution, the mean remaining life of $B$ is just the mean of $B$. $\endgroup$ – Math1000 Jan 23 '20 at 0:59
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The answer has already been provided in the comments. Due to the memorylessness of the exponential distribution, if $A$ fails first, the mean remaining operating life of $B$ at the time of failure of $A$ is the mean operating life of $B$.

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