Reliability Probability problem

Question

What is the Probability that at least one close path is formed from $A$ to $B$ where each switch has a probability of close $p$ and each switch is mutually independent of each other?

Proposed Solution

Let event $A$ be such that

$$\begin{align} A &= \{\text{Current Flows From A to B}\} \\ &= \{ S_0^c S_1^c S_2^c S_3^c, S_0^o S_1^o S_2^c S_3^c , S_0^c S_1^o S_2^c S_3^c, S_0^o S_1^c S_2^c S_3^c, S_0^c S_1^c S_2^o S_3^c \} \\ \end{align}$$

where $S_0^c$ or $S_0^o$ shows the switch $S_0$ Close or Open respectively. Now the probability of $A$ is

$$\begin{align} p[A] &= p^4 + p^2(1-p)^2 + p^3(1-p) + p^3(1-p) + p^3(1-p) \\ &= p^2(1+pq)\\ \end{align}$$

where $q = 1-p$.

Answer 1

I agree with your answer.

$S_3$ must be closed regardless.

Then, the rest of the cases for which at least one of the paths connects are covered by the following:

• $S_2$ is closed; $S_0$ and $S_1$ have any other state (probability $p$)
• $S_0$ and $S_1$ closed, $S_2$ open (probability $p^2q$)

So the total probability is then $p(p + p^2q) = p^2(1+pq)$, which agrees with your answer.

Answer 2

Another way is, to calculate with the converse probability.

Probability, that the system ($S_0,S_1$ and $S_2$) doesn´t work: $(1-p^2) \cdot (1-p)$

So it works with the probabilty of $1-(1-p^2) \cdot (1-p)$

Thus, the whole system work with the probability of $(1-(1-p^2) \cdot (1-p))\cdot p=-p^4+p^3+p^2$

greetings,

calculus

Answer 3

Notice that

$$\begin{align} p[A] &= p^4 + p^2(1-p)^2 + p^3(1-p) + p^3(1-p) + p^3(1-p) \\ &= p^4+3p^3(1-p)+p^2(1-p)^2\\ &= -p^4+p^3+p^2 \end{align}$$

where $q = 1-p$ and $-p^4+p^3+p^2=p^2(1+pq)$ so as others nocited: your calculation for the event probability is correct.