Calculate CDF of system lifetime Consider a system of 4 independent components $A,B,C,D$. The system works if $(A \wedge B) \vee (C \wedge D)$ work. Let $T$ be the random variable describing the system's lifetime, where each component $T_k, k\in\{A,B,C,D\}$ is exponentially distributed with parameter $a$.
I now want to show that $\mathbb{P}(T < t) = (1-e^{2at})^2$, $(t\geq0)$.
I wrote $T$ as the function $max(T_{AB}, T_{CD}) = (T_{AB} \vee T_{CD})$, where $T_{AB} = min(T_{A}, T_{B})$ and $T_{CD} = min(T_{C}, T_{D})$.
Now, to calculate for instance the lifetime of one of the subsystems: $$\mathbb{P}(min(T_{A}, T_{B}) < t) = \mathbb{P}(T_{A} \wedge T_{B} < t) = \mathbb{P}(T_{A} < t,  T_{B} < t) = \mathbb{P}(T_{A}< t)\mathbb{P}(T_{B}< t) = (1-e^{-at})^2$$
How can I now calculate $\mathbb{P}(T < t) = \mathbb{P}(max(T_{AB}, T_{CD}) < t)$?
 A: Your  calculation for the minimum of two components is wrong. 
Your assumption  $$\mathbb{P}(T_{A} \wedge T_{B} < t) = \mathbb{P}(T_{A} < t,  T_{B} < t) $$ is false. It's true for the maximum, so it holds $$\mathbb{P}(T_{A} \vee T_{B} < t) = \mathbb{P}(T_{A} < t,  T_{B} < t) $$
For the minimum you have to calculate:
$$\begin{align} \mathbb{P}(T_{A} \wedge T_{B} < t) &= 1 - \mathbb{P}(T_{A} \wedge T_{B} \ge t) \\&= 1-\mathbb{P}(T_{A} \ge t,  T_{B} \ge t) \\&= 1-\mathbb{P}(T_{A} \ge t)\mathbb{P}(T_{B} \ge t) \\ &= 1-e^{-2at}\end{align} $$
If A,B,C,D are independent components then $T_{AB}$ and $T_{CD}$ are independent and it holds: $$\begin{align} \mathbb{P}(max(T_{AB}, T_{CD}) < t) &= \mathbb{P}(T_{AB} < t,  T_{CB} < t) \\ &= \mathbb{P}(T_{AB} < t)\mathbb{P}(T_{CB} < t) = \left(1-e^{-2at}\right)^2 \end{align}$$
A: Here is an answer that goes into explicit details which might be useful for those who are working through the computations step-by-step and wish to compare.  I hope this helps:  
We can first consider how to compute the probability of the minimum of two RVs which are exponential distributed: 
The min of two random variables can be obtained from the following argument.  Consider the non-negative real line as shown below:
0------t---------->
We are interested in the event where only one of the two RV's is below or equal to the threshold shown as t.  This can be considered a Bernoulli experiment, with the probability of "success" being when an RV has a value below or equal to t, and "failure" otherwise.  
Continuing this reasoning, the probability of having the minimum below the threshold $t$ is   
$P\left(T_{min} \le t \right)= P \left(\text{success} \right)+P \left(\text{success} \right)-P \left(\text{success}\right)^{2}$
The first two terms are for having a single RV below the threshold (there are two ways this can happen), and the subtraction of the third term removes the case where both are below or equal, which is not of interest because it also means that the maximum is below or equal.
Substituing for the expontial distribution for P(success) we have
$P\left(T_{min} \le t \right)= \left(1-e^{-at} \right)+ \left(1-e^{-at} \right)-\left(1-e^{-at}\right)^{2} = 2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}$
Now consider finding the maximum of two random variables:
Using a similar argument to the one above, we get
$P\left(T_{max} \le t \right) = 1-P\left(T_{max} \gt t\right)$
This can be written as:
$P\left(T_{max} \le t \right) = 1-P\left(T_{max} \gt t\right)=1-\left[\left(1-P\left(\text{success}\right) \right) + \left(1-P\left(\text{success}\right) \right) - \left(1-P\left(\text{success}\right) \right)^{2}\right]$
The first two terms indicate the probability of having only a single RV above the threshold and we subtract the case were both are above.  Note that we use $1-P\left(\text{success}\right)$ to indicate these probabilities, because a "success" means having a value below or equal to a threshold, so we are interested in the "failures" meaning that the value is above.
Finding the max of min RVs:
To complete the solution, we combine the above results to obtain:
$P\left(T_{max} \le t \right) = 1-P\left(T_{max} \gt t\right)=1-\left[\left(1-\left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}\right) \right) + \left(1-\left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}\right) \right) - \left(1-\left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}\right) \right)^{2}\right]$
where we have substituted the $P\left(T_{min} \le t \right)$ result for $P\left(\text{success}\right)$.
Simplifying:
$P\left(T_{max} \le t \right)=1-\left[2-4\left(1-e^{-at}\right)+2\left(1-e^{-at}\right)^{2} - \left(1-\left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}\right) \right)^{2}\right]$
$P\left(T_{max} \le t \right)=1-\left[2-4\left(1-e^{-at}\right)+2\left(1-e^{-at}\right)^{2} - \left(1-2\left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}\right) + \left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}\right)^{2}\right)\right]$
$P\left(T_{max} \le t \right)=1-\left[2-4\left(1-e^{-at}\right)+2\left(1-e^{-at}\right)^{2} - 1+2\left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2} \right)- \left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}\right)^{2}\right]$
$P\left(T_{max} \le t \right)=1-\left[1 - \left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}\right)^{2}\right]=\left(2\left(1-e^{-at}\right)-\left(1-e^{-at}\right)^{2}\right)^{2}$
$P\left(T_{max} \le t \right)=\left(2-2e^{-at}-\left(1-2e^{-at}+e^{-2at}\right)\right)^{2}$
$P\left(T_{max} \le t \right)=\left(1-e^{-2at}\right)^{2}$
