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From Mathematical Statistics, 7th ed., Chapter 2, Supplementary Exercise no. 2.163:

Relays used in the construction of electric circuits function properly with probability $0.9$.
Assuming that the circuits operate independently, which of the following circuit designs yields the higher probability that current will flow when the relays are activated? EE

Assuming by "relays" they mean the circles with numbers inside of them (I am not very experienced with Electrical Engineering), I see how this is an exercise meant to test your understanding of conditional probabilities, but I'm having some trouble. For instance:

  • Is it reasonable to assume that, say, circuit $A$ is functioning properly if and only if all of its relays are working?
  • It seems like in circuit $B$, relays $3$ and $4$ are dependent on relays $1$ and $2$, but in circuit $A$ relays $3$ and $4$ are both dependent on $1$ and $2$. Is this likely what they're going for?

It also seems reasonable to say for both circuits that relays $1$ and $2$ are independent, as well as relays $3$ and $4$ are independent.

And from here, I'm not sure where to go. Any help is appreciated!

Jose Avilez
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Mailbox
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    If you want a guess: the whole system works if and only if there is a path from $A$ to $B$. So, for the first one (poorly named $A$ as well), you need at least one of $1,2$ and at least one of $3,4$. For the second, poorly named $B$, you need either both $1,3$ or both $2,4$. But, of course, we really don't know what the writers you are quoting were thinking. – lulu May 20 '24 at 23:33
  • @lulu I see, I was assuming that the system works iff there is a cycle between $A$ and $B$. Is that a valid interpretation as well? – Mailbox May 20 '24 at 23:43
  • I doubt it, but who knows? What properties would you guess your cycle should have? If you can't retrace any edge, then both require all $4$ to be live. Doesn't seem interesting. – lulu May 20 '24 at 23:45
  • @lulu I mean that's interesting to me, lol, I couldn't do that in my head! But your interpretation is probably right, given that the next few questions ask you to find the probability that relays $1$ and $4$ are working. – Mailbox May 20 '24 at 23:48
  • Neither the independence assumption, nor the particular value (0.9) is needed. One may just observe that whenever the circuit B works, the circuit A works as well. So the latter has higher probability of current flowing. – zhoraster May 21 '24 at 19:53
  • @zhoraster Could you elaborate? I don't really see what you mean--the problem states that the circuits are independent, and I don't see any relation between circuits A and B according to the diagram. – Mailbox May 21 '24 at 22:30
  • Circuit B works iff 1 and 3 work or 2 and 4 work. In such case, circuit A works as well. – zhoraster May 22 '24 at 05:52

1 Answers1

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Without getting into too much Electrical Engineering , I have edited the Picture to rename the Circuits $AX$ & $BY$ , & to compete the Ciruits with the "load" & "Power Source" :

EE

Basically , the "load" (Green) is some Device we want to run , like motor or washing machine or lighting , while the "Power Source" (Blue) is what gives Voltage & Current & Energy , like the Utility Connection or Battery or UPS.

The relay is a way to stop the "load" when some bad condition occurs or to start the "load" when some good condition occurs.

Here , relay failure is unwanted.
Exercise is about calculating Circuit failure due to relay failure.

In $AX$ , the "load" will be energized when "1 or 2 are working" and "3 or 4 are working" , to get a complete cycle.
In $BY$ , the "load" will be energized when "1 and 3 are working" or "2 and 4 are working" , to get a complete cycle.

Can you make the calculations now ?
Let me know whether you want me to include the Detailed Calculations here itself.

$AX$ : $2 \times (0.1 \times 0.1) - (0.1 \times 0.1)^2 \approx 0.0199$ Probability of failure.
$BY$ : $(2 \times 0.1 - 0.1 \times 0.1)^2 \approx 0.0361$ Probability of failure.

Probability that Current will flow : $AX$ is higher.

Prem
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    In order to get a complete cycle in $BY$, wouldn't you need either relays 1 and 3 or relays 2 and 4 to be working? If only 1 and 2 were working and 3 and 4 were not working, there wouldn't be an A-B path. – Mailbox May 21 '24 at 13:01
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    Nice Catch , that was a Copy-Paste typo , @Mailbox , never-the-less , that typo Catch indicates that you got the Solution here ! Elaborating the Calculations is not necessary ! – Prem May 21 '24 at 13:41