Results 1 to 3 of 3

Thread: independent events

  1. #1
    Feb 2011

    independent events (solved)

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


    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.


    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 system not working as :
    $\displaystyle \hat{P}=(1-P_{1})(1-P_{2})(1-P_{3})(1-P_{n})$

    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:

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


    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....
    Last edited by jwxie; Jun 11th 2011 at 04:22 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Aug 2006
    Quote Originally Posted by jwxie View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Feb 2011
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Independent events
    Posted in the Statistics Forum
    Replies: 4
    Last Post: Apr 28th 2011, 02:26 PM
  2. Are these events independent?
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Oct 16th 2008, 03:53 PM
  3. Independent events
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Apr 29th 2008, 08:24 AM
  4. Please help me out here about Independent Events
    Posted in the Advanced Statistics Forum
    Replies: 8
    Last Post: Nov 4th 2007, 12:01 PM
  5. Independent events
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: Jul 18th 2007, 08:06 PM

Search tags for this page

Search Tags

/mathhelpforum @mathhelpforum