Results 1 to 2 of 2

Math Help - Probability Chain

  1. #1
    Junior Member
    Joined
    Oct 2010
    Posts
    57

    Probability Chain

    Suppose there is a circuit from A to B.
    The current can flow from 1-2, or from 3-4. (it makes sense to draw a picture)
    Each node (1, 2, 3, 4) has a .9 percent chance of success.
    This means that the system has a .9*.9 + .9*.9 - .9^4 chance of success (.9639)

    The question is this: what is the probability of nodes 1 and 3 working, given that the current is flowing.

    I've been working on this for at least an hour now. I've had a few methods that I believe should work, but none of them are giving me the correct answer of .916.


    On the surface, the problem sounds simple.
    I tried finding the probability of node 1 working, given the system working and multiplying it by the probability of node 2 working, given the system is working
    (.9 / .9639) * (.9 / .9639), but for some reason I can't discern, this isn't right.

    I've tried using Venn Diagram, and I get a number which I'm convinced must be right (.81 / .9639) * (.81 / .9639) = .706, and that doesn't work

    I tried phrasing the question in formal conditional probability:
    P((1 and 3) given ((1 and 2) or (1 and 4) or (2 and 3) or (2 and 4)))
    and the whole thing cancels out on me, leaving .9 * .9.


    Could anyone shed some insight here? Three different ideas I have, each of which make sense to me, give different answers, and are all wrong.
    Last edited by Relmiw; September 20th 2011 at 04:34 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    May 2010
    Posts
    1,028
    Thanks
    28

    Re: Probability Chain

    show us a diagram please, or tell us which nodes are connected to A and B.

    The spoiler below assumes that nodes 1&3 are connected to A and 2&4 are connected to B, however it does not give the answer you say is correct.. so i may have made an error.

    Spoiler:


    using the baysian formula:

    P(1&3 working|current flows) = P(1&3 working AND current flows) / P(current flows).

    step 1 find P(1&3 working AND current flows)
    this is equal to:
    P(1&3&4 working AND 2 not working) [this is the case that current flows through 3->4 only)
    +P(1&3&2 working AND 4 not working) [this is the case that current flows through 1->2 only)
    +P(1&3&2&4) [this is the case that current flows both branches

    = 0.9^3 *0.1 + 0.9^3 *0.1 + 0.9^4 = 0.8019

    Step 2 find P(current flows).
    You already worked this out: 0.9639

    Answer
    0.8019/0.9639=0.8319






    Remark: There are only 16 possible outcomes so if comopletely stuck you can always write them all down in a contingency table to get the answers you need.
    Last edited by SpringFan25; September 21st 2011 at 03:19 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Markov Chain of random variables from a primitive markov chain
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: October 19th 2011, 08:12 AM
  2. Markov Chain & Transition Probability Matrix
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: August 20th 2011, 08:20 PM
  3. Probability/Markov Chain?
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: August 11th 2009, 06:18 PM
  4. Replies: 2
    Last Post: May 6th 2009, 10:41 PM
  5. Replies: 2
    Last Post: October 28th 2008, 06:32 PM

Search Tags


/mathhelpforum @mathhelpforum