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.

$\displaystyle 1-[(1-P_{1}P_{2})(1-P_{3}P_{4})]=1-[1-P_{3}P_{4}-P_{1}P_{2}+P_{1}P_{2}P_{3}P_{4}]$ does give the following result after multiplying P5

$\displaystyle P_{5}(P_{3}P_{4}+P_{1}P_{2}-P_{1}P_{2}P_{3}P_{4})$.

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, $\displaystyle P(A\cup B)=P(A)P(B)$

Let's concentrate on the first part. We are given a system of series and parallel. Treat $\displaystyle P_{i}$ 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 $\displaystyle P(working)=P_{1}P_{2}P_{n}$

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

For elements in parallel, when $\displaystyle P_{1}$ is not working, the system will work if at least one of the remaining elements, $\displaystyle P_{2}$, $\displaystyle P_{3}$ is (or are) still working.

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

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

and etc.

We denote $\displaystyle \hat{P}$, the probability of the whole systemnot workingas :

$\displaystyle \hat{P}=(1-P_{1})(1-P_{2})(1-P_{3})(1-P_{n})$

Thus, the probability of a working system is $\displaystyle 1-\hat{P}$.

Now, let's solve the system.

$\displaystyle P_{upper-branch} = P_{1}P_{2}P_{3}$

$\displaystyle P_{lower-branch} = P_{4}P_{5}P_{6}$

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

$\displaystyle 1-\hat{P} = 1-(1- P_{1}P_{2}P_{3})(1-P_{4}P_{5}P_{6})$

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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:

$\displaystyle 1-[(1-P_{1}P_{2})(1-P_{3}P_{4})]=1-[1-P_{3}P_{4}-P_{1}P_{2}+P_{1}P_{2}P_{3}P_{4}]$

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

$\displaystyle P_{5}(-P_{3}P_{4}-P_{1}P_{2}+P_{1}P_{2}P_{3}P_{4})$

The solution is, however,

$\displaystyle P_{5}(P_{3}P_{4}+P_{1}P_{2}-P_{1}P_{2}P_{3}P_{4})$.

What is wrong with my logic? For the system working, I know the probability should be $\displaystyle P((p1)\cup (p2)\cup (p3))$, 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....