We study over time the state of a machine obeying the following rules:
If the machine is working at the moment $n-1$, it has probability $\dfrac{1}{ 6}$ to be faulty at the moment $n$.
If the machine is faulty at the moment $n-1$, it has probability $\dfrac{2}{ 3}$ to be faulty at the moment $n$.
Denote $p_n$ the probability that the machine is working at the moment $n$.
Question: Determine a relation between $p_n$ an $p_{n-1}$ for all $n\in \mathbb N^*$.
My try:
Let $W_n$ the event "the machine is working on the moment $n$".
Let $F_n$ the event "the machine is faulty on the moment $n$".
Then the two systems $\Omega_n=\{W_n,F_n\}$ and $\Omega_{n-1}=\{W_{n-1},F_{n-1}\}$ are two complete systems of events and then we have by the formula of total probability:
$$p(F_n)=p(F_n|W_{n-1})p(W_{n-1})+p(F_n|F_{n-1})p(F_{n-1})=\dfrac{1}{6}W_{n-1}+\dfrac{2}{3}F_{n-1}$$ Now since $p(F_n)=1-p(W_n)=1-p_n$ and $F_{n-1}=1-W_{n-1}$ then we get $$p_n=\dfrac{1}{2}p_{n-1}+\dfrac{1}{3}$$
Is my try correct? Thank your for your help!!
We will use fewer symbols, and duplicate your result, which is correct. The machine can be working at time $n$ in two disjoint ways: (i) It was working at time $n-1$, and is still working or (ii) It was not working at time $n-1$, but has been fixed.
The probability of (i) is $(p_{n-1})(5/6)$.
The probability of (ii) is $(1-p_{n-1})(1/3)$.