Probability that current could pass through

[pp123123]

Homework Statement

http://imageshack.com/a/img546/5686/nqpn.jpg
I am looking for the probability of the current passing from the left to right given that there are 5 independent switches of probability p. However I cannot figure out the correct solution.

Homework Equations

The Attempt at a Solution

Case I: Assume Switch E is closed so I have a probability of
P((AB) or (CD))=P(AB)+P(CD)-P(AB and CD)=2p^2-p^4

Case II: Assume Switch E is opened so I have a probability of
P((A or C) and (B or D))=(P(A)+P(C)-P(AC))(P(B)+P(D)-P(BD))=(2p-p^2)^2

so overall probability is (1-p)(2p^2-p^4)+p(2p-p^2)^2 which could be simplified to 2p^5-5p^4+2p^3+2p^2

however some friends of mine told be that it should be p^5-5p^4+2p^3+2p^2. I hope to know what's wrong in my calculation. Thanks!

 

[vela]

I got the same answer you did by enumerating all 32 possibilities, identifying which ones belong in the event, and adding up the corresponding probabilities.

You should also consider what the probability should be if p=1. Your friends' expression says the probability is 0. Does that make sense?

 

[Ray Vickson]

Homework Statement

http://imageshack.com/a/img546/5686/nqpn.jpg
I am looking for the probability of the current passing from the left to right given that there are 5 independent switches of probability p. However I cannot figure out the correct solution.

Homework Equations

The Attempt at a Solution

Case I: Assume Switch E is closed so I have a probability of
P((AB) or (CD))=P(AB)+P(CD)-P(AB and CD)=2p^2-p^4

Case II: Assume Switch E is opened so I have a probability of
P((A or C) and (B or D))=(P(A)+P(C)-P(AC))(P(B)+P(D)-P(BD))=(2p-p^2)^2

so overall probability is (1-p)(2p^2-p^4)+p(2p-p^2)^2 which could be simplified to 2p^5-5p^4+2p^3+2p^2

however some friends of mine told be that it should be p^5-5p^4+2p^3+2p^2. I hope to know what's wrong in my calculation. Thanks!

You have not said whether p is the probability that a switch is open (blocking current) or closed (allowing current). I will assume p = probability the switch is closed, allowing current.

There are only 4 paths (routes) from start S to finish F, namely: R1 = ab, R2 = aed, R3 = cd, R4 = ceb. Current can go from S to F if at least one of R1, R2, R3, R4 is 'closed' (meaning it allows current). The probability that R1 is closed is P(R1) = p^2, since both a and b must be closed. You can get the other route probabilities in a similar way. Then, all you need to do is figure out how to find the probability that at least one route is closed, given all the individual route probabilities, etc.

 

[willem2]

. I hope to know what's wrong in my calculation. Thanks!

Your calculation is correct, except that case 1 is really for switch E open, and case 2 for switch E closed, instead of the other way around.