1. ## independent events (solved)

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.

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

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

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

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

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

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

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

and etc.

We denote $\hat{P}$, the probability of the whole system not working as :
$\hat{P}=(1-P_{1})(1-P_{2})(1-P_{3})(1-P_{n})$

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

Now, let's solve the system.

$P_{upper-branch} = P_{1}P_{2}P_{3}$
$P_{lower-branch} = P_{4}P_{5}P_{6}$

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

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

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

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

The solution is, however,

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

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

2. Originally Posted by jwxie
I will start off with my attempt based on the lecture notes, and then present the solution...
Why not start off by stating the actual problem?
Why are you asking us to guess as to what what you are trying to do?

3. Hi Plato. Thanks for the suggestion.
I thought it would be more convenient to state a sample problem first (it's actually very similar except P5). I will state the problem first, then show my approach next time (I usually do that). The book doesn't say anything about series and parallel, and my instructor had warned us that they are not necessarily eletrical elements (resistors, capacitors), just arbitrary "elements". was afraid that the helpers won't understand "our language" so I started off by stating the approach.

I think I got the right answer, but my carelessness put me through this long post. It took me less than ten seconds to review the sign changes after making the post, but at first I spent two hours thinking I had the wrong approach.