MHB Probability that current could pass through

AI Thread Summary
The discussion focuses on calculating the probability of current passing through a series of independent switches (a, b, c, d, e) with a probability p for each switch. Two cases are considered: one where switch E is closed and another where it is open, leading to different probability expressions. The original calculation yields an overall probability of 2p^5 - 5p^4 + 2p^3 + 2p^2, which is confirmed as correct. The alternative method discussed involves counting the configurations of open and closed switches to arrive at the same result. The conclusion affirms that the initial calculation is accurate, countering claims from peers.
pp123123
Messages
5
Reaction score
0
Hi guys. Here's the problem.View attachment 1981
I need to determine the probability that the current could go from left to right given that there is a probability of p for each independent switch named a,b,c,d and e.

I have tried to solve this but I have no idea which part I am getting off the right track. Here is what I am trying to do.

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. Thanksss!
 

Attachments

  • mathhelp.jpg
    mathhelp.jpg
    6.3 KB · Views: 98
Mathematics news on Phys.org
pp123123 said:
Hi guys. Here's the problem.View attachment 1981
I need to determine the probability that the current could go from left to right given that there is a probability of p for each independent switch named a,b,c,d and e.

I have tried to solve this but I have no idea which part I am getting off the right track. Here is what I am trying to do.

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. Thanksss!
Hi pp123123, and welcome to MHB! Short story: you are right and your friends are wrong.

Longer explanation: Your method is correct. Another (more laborious) way to do the calculation is to count the number of open switches.

If all five switches are on then the current gets through. Probability of this is $p^5$.

There are five ways in which four switches can be on, and the current gets through in each case. Probability here is $5p^4(1-p)$.

There are ten ways in which three switches can be on, and the current gets through in eight of these cases. (The only ways in which it cannot get through is if both of the switches on the left part of the circuit, or both of the switches on the right part of the circuit, are off.) Probability here is $8p^3(1-p)^2$.

There are ten ways in which two switches can be on, and the current gets through in two of these cases. (The only ways in which it can get through is if both of the switches on the upper part of the circuit, or both of the switches on the lower part of the circuit, are on.) Probability here is $2p^2(1-p)^3$.

If only one switch, or none at all, is open, then the current cannot get through.

Thus the overall probability is $p^5 + 5p^4(1-p) + 8p^3(1-p)^2 + 2p^2(1-p)^3$. When you multiply out the brackets, this simplifies to $2p^5-5p^4+2p^3+2p^2$.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top