poisson process of a machine with 2 components

Question

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?

Answer 1

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.

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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))