I will start off with my attempt based on the lecture notes, and then present the solution...

APOLOGY IF YOU ARE READING THIS:

I already found my mistake within a minute of my posting. I didn't distribute the minus sign into the expression. I actually had it right.

does give the following result after multiplying P5

.

I think it's helpful to leave this thread alone, so someone else who needs help for the same problem can come and read.

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Suppose we are given a system as shown in the picture below (the problem is the upper part of the image)

For A,B independent events,

Let's concentrate on the first part. We are given a system of series and parallel. Treat as an arbitrary element. Our goal is to look for"the probability that the whole system is in the working condition".

We divide the problem into two sections: series and parallel.

For series

For elements in series, the

For parallel(refers to the lower left of the image)

For elements in parallel, when is not working, the system will work if at least one of the remaining elements, , is (or are) still working.

If doesn't work, then the P(working | P_{1}^{c}) = (1-P_{1}).

If doesn't work, then the P(working | P_{2}^{c}) = (1-P_{2}).

and etc.

We denote , the probability of the whole systemnot workingas :

Thus, the probability of a working system is .

Now, let's solve the system.

Lower and upper are in parallel, and the P(working) is simply:

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I use the methods above to solve the following problem:

The probability of the closing of the i-th relay in the circuits shown is given by pi , for i = 1; : : : ; 5. If all relays function independently, what is the probability that a current ows between A and B for the respective circuits in the image below?

For current to flow, there must be at least ONE closed path.

three paths are available: (1) when 1,2,5 are closed, and (2) when 3,4,5 are closed, or when (1,2,3,4,5) are all closed.

I used the same approach, separate the problem into series and parallel.

But at the end I got

for the parallel:

note that I got the right answer. But I forgot to distribute the minus sign.

For the remaining P_5.... my initial attempt was to multiple the above expression by P_{5}, which gives

The solution is, however,

.

What is wrong with my logic? For the system working, I know the probability should be , where p1,p2,p3 are the 3 paths....

I know I can solve it using the union, but I want to apply what my instructor taught us during the lecture....

Can someone please help me? Thank you very much. This is a long post....