1
$\begingroup$

First, apologies for the image, I couldn't find a good way to explain without including it.

I have the following circuit: Number on each component represents the probability this component works

Number on each component represents the probability this component works

Assuming all its components are independent, what is the probability that the circuit doesn't work?

Note: for the circuit to work we need at least one line from left to right with all components working

I am posting this for two reasons:

  1. I think my solution is long and I might be potentially missing something to make things easier
  2. I got stuck somewhere while solving

My attempt:

I tried doing $1-P(W)$

Let $W$ be the event that the circuit works. For this, we always need $A$ and work, and then either $B$ OR $C$ AND $D$ OR $E$ AND $F$, mathematically this can be written $P(A \cap (B \cup (C\cap D) \cup (E \cap F)))$, and I am not sure how I could proceed with it, the $\cap$ is easy to do since these are independent events I can just multiply them but I don't know how to go ahead with all this $\cup$, I thought about the general formula of $P(A\cap B \cap C=P(A)+P(B)+P(C)-P(A\cap B) \dots $ but I am not quite sure of it

I also tried doing $P(W')$ directly but it seemed longer.

Any help would be appreciated.

$\endgroup$
4
  • $\begingroup$ To clarify, it looks like you mean the circuit "works" as long as there's a path from the left side to the right side on which all the components "work". $\endgroup$ – Karl Feb 13 at 8:30
  • $\begingroup$ That is correct $\endgroup$ – Sergio Feb 13 at 8:33
  • $\begingroup$ added this as well as some edits for better clarity. $\endgroup$ – Sergio Feb 13 at 8:37
  • $\begingroup$ The basic rules are: If two circuit elements are in series (i.e. one after the other) then one multiplies the individual success probabilities to get the combined success probability. If they're in parallel (i.e. start at common vertex and end at common vertex) replace "success" with "failure" in the above. $\endgroup$ – Semiclassical Feb 13 at 8:53
1
$\begingroup$

Let's write some general rule for this kind of probability circuits. Assume for simplicity that the events are all of the kind "work".

$(G1) \qquad $ If you have two or more independent events $C_1, C_2, \dots, C_n$ that are "in series" then you can collapse them all togheter in an event, say $C'$, that has probability of success $$P(C') = \prod_{i=1,\dots, n} P(C_i)$$

$(G2) \qquad$ If you have two or more independent events $C_1, C_2, \dots, C_n$ "in parallel" then you can collapse them all togheter in an event, say $C'$, that has probability of success $$P(C') = 1- \prod_{i=1,\dots, n} (1-P(C_i))$$

A basic proof of $(G2)$ can be seen like this: The probability that a parallel circuit will work is exactly the probability that at least one component will work. Using the "at least one" principle we can conclude that the probability that the whole circuit will work is $1$ minus the probability that all the component will not work simultaneously.

In your probem you have the couples $C,D$ and $E,F$ in series, so they can be combined in two events $C'$ and $E'$ that have probability of success $$P(C') = P(C)P(D) = 0.525 \hspace{1.5cm} P(E') = P(E)P(F) = 0.68$$

Now you have $B$, $C'$, $E'$ in parallel. Theese can be sees as a unique event $B'$ that has probability of success $$P(B') = 1 - (1-P(B))(1-P(C'))(1-P(E')) \sim 0.94$$

Finally we arrived at two events $A$ and $A'$ in series and we can conclude that the probability that the event $W$ "the whole circuit works" is:

$$P(W) = P(A)P(B') \sim 0.56$$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.