Consider a system with three identical components and their fault rate is exponentially distributed with $\lambda > 0$. The system needs all three components to function (series system). If you have one extra component that can replace one faulty component (with no install time) what is the probability that the system survives up to time $T$. The failures for each components is independent.
I tought the probability is the complement of event that two out of three components fail during time $T$, because the extra components allows one component to fail during $t\in[0,T]$. If the components are labeled $C_1,C_2,C_2$ possible combinations for failure events would be $F=C_1 C_2 + C_1 C_3 + C_2 C_3$ which would result into the system not working. This results into total probability of
$$P(F) = P(C_1)P(C_2)+P(C_1)P(C_3)+P(C_2)P(C_3)-2\cdot P(C_1) P(C_2) P(C_3)$$ And when we consider the time the failure would be same for each component given by the cumulative distribution function: $P(t \leq T \text{ for } C_1,C_2,C_3=1) = 1-e^{-\lambda T}$ \begin{align*}P(\text{survives to }T) &= 1 - P(\text{two components fail in } t\in [0,T])\\ &=1-3\cdot(1-e^{-\lambda T})^2 + 2\cdot (1-e^{-\lambda T} )^3\end{align*} Does this seem correct?
EDIT: Using what @joriki said. Probability that twice in a row one out of three components fail.
\begin{align*} P(2\times \text{1 out of 3 fail}) &= P(C_1 \overline C_2 \overline C_3 \cup \overline C_1 C_2 \overline C_3 \cup \overline C_1 \overline C_2 C_3)^2 \\ &=((1-e^{-\lambda T})e^{-2\lambda T} + (1-e^{-\lambda T})e^{-2\lambda T} + (1-e^{-\lambda T})e^{-2\lambda T})^2\\ &= 3^2 (1-e^{-\lambda T})^2 e^{-4\lambda T}\\ P(\text{survives to }T) &= 1- 9 (1-e^{-\lambda T})^2 e^{-4\lambda T} \end{align*} Like this?