poisson process of a machine with 2 components I´m having trouble with a question for my Statistics class. Say a system works using 2 components, and stops working whenever one of those 2 components break. Component A fails on average 1 time every 400 hours, and component B fails on average 1 time every 600 hours. What is the probability that the system stops working after the first 800 hours? And what is the probability that the system stops working after the first 800 hours if, after an upgrade, it only stops working when both components fail?
For the first time, I worked considering a gamma distribution, and estimated the probability for each component to fail after 800 hours. For A, P(X >= 800) given alpha=1 and beta=400, is 0,1353. And for B, P(X >= 800) given alpha=1 and beta=600, is 0,2231.
So, given that each component is independent, I estimated the probability that the system fails as P(AUB) = 0,1353 + O,2232 - (0,1353*0,2232) = 0,328
And for the last question, I estimated that probability as P(AandB) = (0,1353*0,2232) = 0,03.
However, I think I'm doing it wrong. Could someone help me with this question?
 A: Outline: 
$X$ is time to failure of A;  $X \sim 
\mathsf{Exp}(\mathrm{rate}=\lambda_a = 1/400).$
$Y$ is time to failure of B;  $Y \sim 
\mathsf{Exp}(\mathrm{rate}=\lambda_b = 1/600).$
Let $V = \min(X,Y)$ be the time to first failure; one can show that
$V \sim 
\mathsf{Exp}(\mathrm{rate}=\lambda_a + \lambda_b = 1/240).$
The method of proof is:
$$1 - F_V(v) = P(V > v) = P(X >v, Y > v)\\ = P(X > v)P(Y > v) = e^{-\lambda_a v} e^{-\lambda_b v} =\cdots.$$
You seek $P(V > 800) = 1 - P(V \le 800) = 1 - F_V(800) = 0.0357.$
Using R statistical software, this can be evaluated as
shown below, but you can also evaluate exponentials on a
calculator.
1 - pexp(800, 1/240)
[1] 0.03567399

The part after the upgrade requires $W = \max(X,Y).$
The CDF of $W$ can be found as follows:
$$F_W(w) = P(W \le w) = P(X \le w, Y \le w)\\ = P(X\le w)P(Y\le w) \cdots.$$
However, the distribution of $W$ is not another exponential
distribution. But $F_W(w)$ can still be used; it's just a bit
messier.

A simulation of a million 2-component systems in R gives approximate numerical solutions
and makes it easy to show histograms of the distributions
of $V$ and $W$ which approximate the shapes of their two
distributions.
set.seed(2020)
m = 10^6; lam.a = 1/400; lam.b = 1/600
x = rexp(m, lam.a);  y = rexp(m, lam.b)
v = pmin(x,y);  w = pmax(x,y)
mean(v)
[1] 240.0647   # aprx E(V) = 240
mean(v > 800)  
[1] 0.035672   # aprx P(V > 800) = 0.0357
mean(w)
[1] 760.2749   # aprx E(W)
mean(w > 800)
[1] 0.363614   # aprx P(W > 800)

par(mfrow=c(2,1))
hist(v, prob=T, br=40, col="skyblue2", 
     main="Simulated Dist'n of V")
  curve(dexp(x, 1/240), add=T, col="red", lwd=2)
  abline(v = 800, lty="dotted")
hist(w, prob=T, br=40, col="skyblue2", 
    main="Simulated Dist'n of W")
  abline(v = 800, lty="dotted")
par(mfrow=c(1,1))


