Show $P(T < t) =(1-e^{-2\alpha t})^{2}$

A system consists of four components that are similar and work independently, say $$A,B,C,D$$. To operate properly, it is necessary that $$(A$$ and $$B)$$ or $$(C$$ and $$D)$$ are functioning properly.

Define $$T$$ as the failure time of the complete system, and $$T_{k}$$ as the failure time of component $$k \in \{A,B,C,D\}$$, and suppose that $$T_{k}$$ ~ $$\exp{(\alpha)}$$.

Show that $$P(T < t) =(1-e^{-2\alpha t})^{2}$$

My ideas:

Since we have been given the distribution of $$T_{k}'$$s, we will have to write $$T$$ in terms of $$T_{A},T_{B},T_{C},T_{D}$$

From @antkam's idea, I've got to the following:

Assuming there are only $$T_{A},T_{B}$$, it follows $$T=\min{T_{A},T_{B}}$$

$$P(T< t)=P(\min{T_{A},T_{B}}

• Hint: consider just the (A and B) part of the system. This fails at time $\min(T_A, T_B)$, i.e. the minimum of two exponetial random variables. Do you know how to calculate that? If not, see: en.wikipedia.org/wiki/… – antkam Feb 11 at 18:11
• oh, you have multiple misunderstandings then... first, when a component fails, it does NOT recover. it remains failed for all time afterwards. so e.g. if A fails at $t=3$, it remains failed for all $t \ge 3$. the exponential $exp(\alpha)$ describes that one moment when A starts to fail. second, if A fails $t \ge 3$ and B fails $t \ge 5$, then the (A and B) system has failed at time $t \ge 3$ because it requires both A and B to function. – antkam Feb 12 at 16:20
• so lets solve the subsystem E = (A and B) first. we have $T_E = \min (T_A, T_B)$ because whichever of A/B fails earlier, that's when E fails. (and remember, nobody ever recovers.) now $T_A, T_B$ are both exponential $exp(\alpha)$. look into my earlier wiki link to find the distribution of $T_E$. can you answer the question: what is $P(T_E < t)$? – antkam Feb 12 at 16:22
• so you're almost done... F = (C and D) has the same $P(T_F < t) = 1 - exp(-2\alpha t)$. Now back to the whole system of (E or F), what is $P(T < t)$? – antkam Feb 12 at 16:52
• Thanks for your help, I think my confusion simply stemmed from the fact that $T$ shows time at which the system fails rather than time duration of the system failure – SABOY Feb 12 at 17:02