Conditional probability in component network

Question

I am a little stuck on this problem involving the below reliability network:

The probability that any given component works is independent with probability $p$, and fails with probability $q=1-p$. I am trying to find the probability that components $1$ and $5$ work given that the system works.

Starting with the probability the system works I get the probability that components $1$ and $2$ work is $p^2$, the probability that components $4$ and $5$ work is $1-q^2$. Therefore the probability that components $3$, $4$, and $5$ work is $p(1-q^2)$, hence:

$$P(\textrm{System Works})=1-(1-p^2)(1-p(1-q^2))$$

This can be simplified but I do not think it is too important. I then try to find the probability that $1$, $5$, and the system works. I am struggling on this part, it seems like there are many combinations where this condition is satisfied, so I was wondering if there is a better way to consider all the events which the system works as well as components $1$ and $5$?

Just to clarify what I mean by a reliability network. The probability that the system works is a path getting from the right to left side (or vice versa) passing through components that work with probability $p$. You cannot pass through a component if it fails (probability $1-p$ defined to be $q$).

Answer 1

\begin{eqnarray*} \mbox{Let } && S = \mbox{System works} \\ && W = \mbox{1 and 5 both work.} \end{eqnarray*}

With the obvious notation, and using distributivity rules: \begin{eqnarray*} S &=& (3 \cap (4 \cup 5)) \;\cup\; (1 \cap 2) \\ &=& (3 \cap 4) \;\cup\; (3 \cap 5) \;\cup\; (1 \cap 2). \\ W &=& 1 \cap 5. \\ \therefore S \cap W &=& \left[(3 \cap 4) \;\cup\; (3 \cap 5) \;\cup\; (1 \cap 2)\right] \;\cap\; (1 \cap 5) \\ &=& (1 \cap 3 \cap 4 \cap 5) \;\cup\; (1 \cap 3 \cap 5) \;\cup\; (1 \cap 2 \cap 5) \\ &=& (1 \cap 3 \cap 5) \;\cup\; (1 \cap 2 \cap 5). \\ \therefore P(S \cap W) &=& P(1 \cap 3 \cap 5) + P(1 \cap 2 \cap 5) - P(1 \cap 2 \cap 3 \cap 5) \quad\mbox{by inclusion-exclusion} \\ &=& 2p^3 - p^4. \end{eqnarray*}

Answer 2

I would use chain rule for:

$P$(System works,One ok,Five ok)=$P$(System works|One ok,Five ok)$P$(One ok,Five ok)

Assuming one ok and five ok:

• one can be eliminated from the network (passes always through).
• similarly you can eliminate five - there is just a straight line
• finally, if you pass five all the time, you can eliminate four: you can use always the line where was five
• the remaining nodes are two and three, in the same arrangement as four and five in the original picture

Therefore we get:

$P$(System works|One ok,Five ok) = $1-q^2$

From the formulation and independence, we get

$P$(One ok,Five ok) = $p^2$

Thus $P$(System works,One ok,Five ok) = $p^2(1-q^2)$