# The mean remaining operation lifetime of a parallel system

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 $$(X_{A} < X_{B})$$?

• By the memoryless property of the exponential distribution, the mean remaining life of $B$ is just the mean of $B$. – Math1000 Jan 23 '20 at 0:59

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 $$B$$.