Probability of survival for system

Question

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?

Answer 1

The probability that three components are still working after time $t$ is $\mathrm e^{-3\lambda t}$. The probability density for the first of them to fail at time $t$ is the negative derivative of this, $3\lambda\mathrm e^{-3\lambda t}$. The probability for two such failures to occur within time $T$ is

$$ \int_0^T3\lambda\mathrm e^{-3\lambda t}\left(1-\mathrm e^{-3\lambda(T-t)}\right)\mathrm dt=1-\mathrm e^{-3\lambda T}-3\lambda T\mathrm e^{-3\lambda T}\;, $$

where the first factor in the integrand is the density for the first failure to occur at $t$ and the second factor is the probability for the second failure to occur before the remaining time $T-t$.

Answer 2

I have a little trouble wrapping my head around this one, too, so I offer a somewhat questioning answer. The exponential distribution in question is $\lambda e^{-\lambda t}$ which means that the probability of a given component failing by time $t_1$ is $$\int_0^{t_1}\lambda e^{-\lambda t}dt=1-e^{-\lambda t_1}$$ That means that the probability of surviving up to time $t_1$ is $e^{-\lambda t_1}$, so the probability of failing between time $t_1$ and $t_1+dt$ is $\lambda e^{-\lambda t_1}dt$, which in turn means that the probability of failing between $t_1$ and $t_1+dt$ given survival up to $t_1$ is $\lambda dt$. So we can set up the probability of surviving up to time $t$, failing between time $t$ and $t+dt$ and then surviving up to time $T$ as $$\left(e^{-\lambda t}\right)^3(3\lambda dt)\left(e^{-\lambda(T-t)}\right)^3$$ Where the exponentials are cubed because we require joint survival of all $3$ components and the infinitesimal probabilities of failure for the individual components add up to $3\lambda dt$. Then we can add up the probabilities over the entire time interval to get the probability of one failure $$P_1=\int_0^T3\lambda e^{-3\lambda T}dt=3\lambda Te^{-3\lambda T}$$ The probability of no failures is simply $P_2=e^{-3\Lambda T}$, so the probability of survival with at most one failure is $P_1+P_2=(1+3\lambda T)e^{-3\lambda T}$. As I said, I'm not $100\%$ sure about this approach so I would appreciate a little constructive criticism.