1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability Q about parallel components

  1. Dec 1, 2012 #1

    CAF123

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    A parallel system functions whenever at least one of its components works. Consider a parallel system of ##n## components, and suppose that each component works independently with prob 1/2. FInd the conditional prob that component 1 works given that the system is functioning.

    3. The attempt at a solution
    Let E be the event that compt 1 works and F the event that the system functions.
    P(E|F) = P(EF)/P(F) = P(F|E)P(E)/P(F). I have the numerator correct, I just have a subtle question about the denominator.
    When I see the words 'at least', I always jump to the conclusion of doing 1-P(less than 'at least'). Is this a normal practice?
    So I get the correct answer for the denominator using this way. However, I also considered finding the denominator by conditioning on the component that works. So I said;
    Let ##G_i## be the event that compt i works. Then $$P(F) = P(\cup_{i=1}^{n} F|G_i) = P(F|G_1)P(G_1) + P(F|G_2)P(G_2) +...+ P(F|G_n)P(G_n)$$. Correct me if I am wrong, but I believe this method fails because this assumes that the system functions when only one component works? In general, a union is where we want only one event to happen out of,say, n possibilities? Thanks!
     
    Last edited: Dec 1, 2012
  2. jcsd
  3. Dec 1, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If E occurs then F occurs (since at least component 1 is functioning); that is, E is a subset of F, so EF = E, giving P(E|F) = P(E)/P(F). Of course, P(F) = 1 - P{all broken}, just as you said; and yes, that is often the best way to do such computations and is a completely normal practice.

    To get the denominator the hard way, you need to compute
    [tex] P(F) = P\{\cup_{i=1}^n G_i \}.[/tex] For example, if n = 3 this would be given by
    [tex] P(F) = P(G_1) + P(G_2) - P(G_1 G_2) = 1/2 + 1/2 - 1/4 = 3/4.[/tex] For larger values of n this gets more complicated, but you can appeal to general versions of the inclusion/exclusion principle as found, eg., in Feller, "An Introduction to Probability Theory and its Applications", Vol. I, Wiley (1968)----my very favorite probability book---and apply the formulas in Chapter IV. Specifically, if
    [tex] S_1 = \sum_i P(G_i), \; S_2 = \sum_{i,j:i < j}P(G_i G_j), \;
    S_3 = \sum_{i,j,k: i < j < k} P(G_i G_j G_k),\\
    \;\;\;\;\; \vdots\\
    S_n = P(G_1 G_2 \cdots G_n),[/tex]
    then the probability that at least r of the events G_i occur is
    [tex] P_r = S_r - {r \choose r-1} S_{r+1} + {r+1 \choose r-1} S_{r+2}
    - {r+2 \choose r-1} S_{r+3} + \cdots \pm {n-1 \choose r-1} S_n. [/tex]
    In your case you can simplify this, using the fact that
    [tex] S_m = {n \choose m} 1/2^m,[/tex] because S_m is just the number of m-tuples chosen from 1,2,...,n, multiplied by a probability that is the same for any m-tuple. It simplifies even more because you only need the case r = 1, so
    [tex]\text{answer } = S_1 - S_2 + S_3 - \cdots \pm S_n,[/tex] and if you write this out in more detail something very nice happens.
     
    Last edited: Dec 1, 2012
  4. Dec 1, 2012 #3

    CAF123

    User Avatar
    Gold Member

    How does this second line follow from the first? Is what I wrote incorrect?Thanks
    Also, in relation to what this union actually means - does it represent the probability of one of the events happening?
     
    Last edited: Dec 1, 2012
  5. Dec 1, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Yes, what you wrote not only makes no sense (look at it carefully!), but even if re-written properly would still be wrong. In general, an equation of the form
    [tex]P(F) = \sum_i P(F|G_i) P(G_i)[/tex]
    is only true if the G_i are disjoint and their union is the whole space. In your problem the G_i are not disjoint, so what you wrote is false.

    What I wrote in the second line is discussed in every introductory probability textbook; it is called the inclusion-exclusion principle. See, eg.
    http://www.saylor.org/site/wp-conte...ia_Inclusion-Exclusion-Principle_6.8.2012.pdf
    or
    http://kaharris.org/teaching/425/Lectures/lec7.pdf
     
  6. Dec 2, 2012 #5

    CAF123

    User Avatar
    Gold Member

    I am still struggling to see why a union calculation would be appropriate here. It is my understanding that a union of events ##G_i## would give the probability of only one of the events ##G_i## occuring, ie giving the prob that only one of the components are working. However, we want that the prob that at least one of the events is occuring so from that union of events, any number of the ##G_i## could occur to satisfy the system functionality, not just one. thanks
     
  7. Dec 2, 2012 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I don't know where your "understanding" about a union comes from---it is flat out wrong!
    "A union B" = "A or B" = "at least one of A or B". If you want the set for "either A or B but not both" you need the set ##(A-B) \cup (B-A).##

    Of course, if A and B are *disjoint* sets then ---- and ONLY then --- will we have that the union is the same as 'one or the other but not both'.

    All this is made clear from the Venn diagrams in the links I provided in my previous response. Did you take the trouble to read them?

    Finally, to see the difference, take two general sets A and B. Let C = {at least one of A or B occur} and D = {exactly one of A or B occur}. Their probabilities are generally different:
    [tex] P(C) = P(A \cup B) = P(A) + P(B) - P(AB)\\
    P(D) = P[(A-B) \cup (B-A)] = P(A) + P(B) - 2P(AB).[/tex]
    Just draw a Venn diagram to see this.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Probability Q about parallel components
  1. Simple probability Q (Replies: 14)

Loading...