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 $\mathbb{P}(T < t) = \mathbb{P}(max(T_{AB}, T_{CD}) < t)$?
Best Answer
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: $$\begin{align} \mathbb{P}(max(T_{AB}, T_{CD}) < t) &= \mathbb{P}(T_{AB} < t, T_{CB} < t) \\ &= \mathbb{P}(T_{AB} < t)\mathbb{P}(T_{CB} < t) = \left(1-e^{-2at}\right)^2 \end{align}$$