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 $T=\min{T_{A},T_{B}}$

$P(T< t)=P(\min{T_{A},T_{B}}<t)=1-P(\min{T_{A},T_{B}}\geq t)=1-P(T_{A}\geq t,T_{B}\geq t)=1-P(T_{A}\geq t)^{2}=1-(\exp(-\alpha t))^{2}=1-\exp(-2\alpha t)$

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    $\begingroup$ 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/… $\endgroup$ – antkam Feb 11 at 18:11
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    $\begingroup$ 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. $\endgroup$ – antkam Feb 12 at 16:20
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    $\begingroup$ 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 $P(T_E < t)$? $\endgroup$ – antkam Feb 12 at 16:22
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    $\begingroup$ 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 $P(T < t)$? $\endgroup$ – antkam Feb 12 at 16:52
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    $\begingroup$ 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 $\endgroup$ – SABOY Feb 12 at 17:02

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