Let us consider a machine with two parts X and Y. The time before part X goes out of service has an Exponential distribution with $\lambda = 1$ and the time before part Y goes out of service has an Exponential distribution with $\lambda = 0.05$ (its life expectation is 20 time units). When the part X breaks it is replaced in 1 time unit and the whole machine is not working during this time (hence the part Y is not ageing while the part X is being replaced). When part Y goes out of service, the whole machine is considered dead. What is the expectation of the random variable T, which stands for the time before death of the whole machine?
I have tried to consider conditioning on the lifespan of the part Y (I denoted this random variable T_Y) and on the number of times the part X breaks before the part Y goes out of service (N). $$ E(T) = \sum_{k=0}^{\infty}\int_{0}^{\infty}E(T|T_Y, N)P(N=k|T_Y)P(T_Y=t_{y})dt_{y} $$ But I am unsure of the formula. Also I would say that $E(T|T_{Y}, N) = t_{y} + k$ but I don't see why exactly. And what would be $ P(N=k|T_{y})$?
Since the exponential distribution is memoryless, each time part $X$ is replaced, we have a new Bernoulli trial, where success means that part $Y$ fails before part $X.$ From the distribution of the difference of two independent exponentially-distributed random variables, (see this question) we calculate that the probability of success is $$p={.05\over1.05}$$ and we know that the expected number of trials until success is $${1\over p}=21$$
That means, that on average, part $X$ fails $20$ times before part $Y$ fails, so $$E(T)=20+E(Y)=20+20=40$$
just as we would have guessed.