What is the probability that current flows from L to R for the circuit shown below??

[arupray]

Can someone help me on the problem:

Assume the probability of each relay being closed is 'p' and that each relay is open or closed independently of other relay. What is the probability that current flows from L to R? See the circuit diagram below. The relays are just like switches. If closed current flows else if open current does not flow.

 

This problem is from the book of Meyer : introduction to probability and stastical applications.

 

 

[studiot]

What have you done towards this question so far?

 

Since you have posted in mathematics, I assume it is the electrical connection that is causing difficulty so to translate from electrical switching circuits to mathematics

 

Parallel connections are equivalent to the mathematical OR or the Union of sets in a venn diagram

 

Series connections are equivalent to the mathematical AND or Intersection of sets in a venn diagram

 

So all the switches can be replaced by their probabilities plus the appropriate connective AND / OR

in a probability tree diagram.

 

These can then be replaced by the mathematical operations + for OR and x for AND operating on the probabilities.

 

For example

Contacts 1 & 3 are connected in parallel

 

The parallel connection containing {1 & 3} is in series with contact 2.

 

So that branch has equation

 

{p OR p} AND {p}

 

{p + p} x {p} = 2p²

Edited August 18, 2015 by studiot

[John Cuthber]

There are no relays in that circuit .

[studiot]

I think it's called artistic licence, John.

 

All mathematicians have it.

 

Edited August 18, 2015 by studiot

[arupray]

Yes i agree the to the concepts of AND / OR . but in that way my answer is 3P^2+P.

BUT THE ANSWER PROVIDED BY THE AUTHOR IS

P+3P^2-4P^3-P^4+3P^5-P^6

[Strange]

I get a different answer from both of you!

 

I don't understand how there can be negative values in there ...

[studiot]

Yes i agree the to the concepts of AND / OR . but in that way my answer is 3P^2+P.

BUT THE ANSWER PROVIDED BY THE AUTHOR IS

P+3P^2-4P^3-P^4+3P^5-P^6

 

Yes you are correct there is more to it than I first thought.

 

If q is the probability of failure of an element (no circuit connection) then the probability of success (a valid connection) is

 

For a series connection of n elements.

 

[math]{P_{success}} = \prod\nolimits_1^n {\left( {1 - q} \right)} [/math]

 

 

and for a parallel connection of n elements

 

[math]{P_{success}} = 1 - \prod\nolimits_1^n {\left( q \right)} [/math]

 

My original thoughts led to

[math]{P_{success}} = p + 3{p^2}[/math]

 

Which is clearly wrong as substituting p =1 shows.

 

Meyer had

 

[math]{P_{success}} = p + 3{p^2} - {p^3} - {p^4} + 3{p^5} - {p^6}[/math]

 

I did not get this when substituting q = 1-p in the above but got

 

[math]{P_{success}} = 3p - 4{p^2} + {p^3} + 4{p^4} - 4{p^5} + {p^6}[/math]

 

I expect Meyer is right and I made one of my silly arithmetic bobos however both my expression and Meyer's leads to a credible distribution running from 0 to 1 as expected as in the following spreadsheet.

 

 

 

Edited August 19, 2015 by studiot

[John Cuthber]

A current can flow through one of three paths.

Via "4" with probability P

Via "5" and "6" with probability P²

or via "2" and either "1" or "3" with probability P X 2P

Adding gives

P + 3P²

 

so I agree with Arupray.

I also agree with Strange's comment "I don't understand how there can be negative values in there ..."

[studiot]

Think about it

 

if p = 1 then p + 3p² = 4.

 

So you are saying there is a probability of 4 of a connection if every switch is closed.

 

This is clearly wrong and what the first column of the spreadsheet was all about.

 

 

I was wrong in my initial statement about the parallel situation; it works like this.

 

If p_i = probability for switch i closed q_i the chance of switch i open

 

and

 

If P = probability for network connection Q the chance for network disconnection

 

 

then

 

p_i + q_i = 1

 

and also

 

P + Q = 1

 

For a series situation

 

1)Any switch open will loose connectivity

2)All switches have to be closed for connection.

 

So the chance of network connection depends individually on every switch

 

P = p₁ x p₂ x p₃ x ..........p_i

 

 

 

For a parallel situation the obverse situation occurs.

 

1)Any switch closed makes continuity.

2)All switches have to be open to loose connectivity

 

This time we calculate the chance of loosing network connectivity since it involves all switches

 

Q = q₁ x q₂ x q₃ x ..........q_i

 

But we want the chance of the network connected, not the chance of it disconnected so

 

P = 1 - Q = 1- {q₁ x q₂ x q₃ x ..........q_i}

 

We can't use case 1) for parallel probability becase the probability for (say) switch 1 being closed is p but this includes situations where switch 2 etc is also closed and cases where it is not. so we would be counting it twice if we simply totted up the probabilities in the parallel branches without subtracting the cases where the connection is made anyway. This is where the negatives come in.

[John Cuthber]

Think about it

 

if p = 1 then p + 3p² = 4.

 

So you are saying there is a probability of 4 of a connection if every switch is closed.

 

This is clearly wrong and what the first column of the spreadsheet was all about.

 

 

I was wrong in my initial statement about the parallel situation; it works like this.

 

If p_i = probability for switch i closed q_i the chance of switch i open

 

and

 

If P = probability for network connection Q the chance for network disconnection

 

 

then

 

p_i + q_i = 1

 

and also

 

P + Q = 1

 

For a series situation

 

1)Any switch open will loose connectivity

2)All switches have to be closed for connection.

 

So the chance of network connection depends individually on every switch

 

P = p₁ x p₂ x p₃ x ..........p_i

 

 

 

For a parallel situation the obverse situation occurs.

 

1)Any switch closed makes continuity.

2)All switches have to be open to loose connectivity

 

This time we calculate the chance of loosing network connectivity since it involves all switches

 

Q = q₁ x q₂ x q₃ x ..........q_i

 

But we want the chance of the network connected, not the chance of it disconnected so

 

P = 1 - Q = 1- {q₁ x q₂ x q₃ x ..........q_i}

 

We can't use case 1) for parallel probability becase the probability for (say) switch 1 being closed is p but this includes situations where switch 2 etc is also closed and cases where it is not. so we would be counting it twice if we simply totted up the probabilities in the parallel branches without subtracting the cases where the connection is made anyway. This is where the negatives come in.

Oops

[studiot]

I hope I explained it well enough, please let me know if not, it is difficult to put the idea into words.

[John Cuthber]

you did a fine job of explaining that I'm a twit.

[arupray]

Excellent explanation. One final query the answer you (studiot) provide and that provided by Meyer are different, although both probability sum up to 1. Which one is the actual answer? Also could you show a more detailed working leading to the answer, using the formulas you have given.

My answer is matching that of Meyer. I have used the concept used by Studiot.

P+3P^2-4P^3-P^4+3P^5-P^6

Edited August 21, 2015 by arupray

[studiot]

Yes, as I thought I had misplaced a 1 in the arithmetic somewhere.

Upon doing it more carefully I find Meyer is unsuprisingly correct.

 

I have attached the solution but a few words might may be appropriate as I have adopted a format and notation that helps keep order with this type of problem.

 

The posh mathematical term for the connection is ' the event' and for the open circuit is 'the complementary event'

 

A less posh term is success for the event and failure for the complementary event.

 

These terms refer to the entire network ( between nodes L and R in this case) but reappear in subsections of the network as 'elemental events'

 

I have used capital letters to denote the whole network or major subsections and lower case to denote elements.

 

So S is success for the whole network and F is failure.

 

S₁, S₂, S₃ and F₁, F₂, F₃ refer to the three main branches

 

s_i and f_i refer to the individual switches 1 through 6.

 

So we have the basic relation S + F = S_i + F_i = s_i + f_i = 1

 

and we use this and apply the assigned probabilities p_i ( all the same = p in this case)

To transform back and fore between success and failure as needed.

 

 

[arupray]

Thank you Studiot. Meyer has solved problems like this of less complexity using Venn Diagrams. But you have a more fundamental concept for. As Venn Diagrams will become clumsy for 4 or more branches of the network. Thank You once again.