Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Cant get Venn diagram to agree with Demorgan's Law

  1. Oct 10, 2016 #1
    I was trying to solve a problem about calculating the probability of failures of a communications link with several nodes, and figured I need to use Demorgan's law since it would involve non mutually exclusive events. So I tried to go back to the basics to make sure I understand the axioms of probability and prove to my self that this is correct using Venn diagrams, but I can't make the numbers come out right trying to apply Demorgan's Law. Here is the example:

    P(A) = .3
    P(B) = .2

    P(A∪B) = P(A) + P(B) - P(A∩B)
    P(A∪B) = .3 + .2 - .05 = .45

    But if I try to apply Demorgans's I don't get what I expect.

    P(A∪B)' = P(A' ∩ B')
    P(A∪B)' = (1-P(A))(1-P(B))
    P(A∪B)' = (1-.3)(1-.2)
    P(A∪B)' = (.7)(.8) = .56

    but clearly P(A∪B)' = .55

    so where am i going wrong? All parts of my Venn diagram add up to 1. Also I believe that the identity P(A∩B) = P(A)P(B) but this doesn't seem to be correct either as .3 x .2 ≠ .05. So I guess i'm misapplying the use of the Venn diagram but I'm not sure how. Can someone please explain what I'm doing wrong?


    Thanks in advance.
  2. jcsd
  3. Oct 10, 2016 #2


    User Avatar
    Science Advisor

    The latter equality only holds if A and B are independent, i.e. if P(A ∩ B) = P(A)P(B) (which implies that also A' and B' are independent), but that is not the case here.
  4. Oct 10, 2016 #3


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    Where does the .05 come from, then ?
  5. Oct 10, 2016 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Your second step is not correct. [itex]P(A' \cap B')[/itex] is only equal to [itex](1-P(A))(1-P(B))[/itex] if [itex]A[/itex] and [itex]B[/itex] are mutually exclusive, that is, if [itex]P(A \cap B) = 0[/itex]
  6. Oct 10, 2016 #5


    User Avatar
    Science Advisor

    Independent, not mutually exclusive. If we change P(A ∩ B) to 0.06, then the formula holds, but A and B are not mutually exclusive.
  7. Oct 10, 2016 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    Right, my mistake.
  8. Oct 11, 2016 #7
    Ok. So can you only use Demorgan's for independent relations? If not I don't see how you can get P(A' ∩ B') if I can't use P(A' ∩ B') = (1 - P(A))(1-P(B)) .
  9. Oct 11, 2016 #8

    Stephen Tashi

    User Avatar
    Science Advisor

    No, DeMorgan's laws aren't restricted to independent events.

    The diagram can be partitioned into disjoint sets labeled according to whether they are in or not-in each of the named sets. This is like a coordinate system. So the distinct areas of the diagram have "coordinates" (A,B), (A,B'),(A',B), (A'B').

    If you want the probability of the set ##A'\cap B'## (which corresponds to (A'B') ) take 1 minus the sum of the probabilities of the other sets in the partition. So ##p(A'\cap B') = 1 - P(A'\cap B) - P(A\cap B') - P(A\cap B)##.
  10. Oct 11, 2016 #9
    Anyways I started this discussion because I was trying to figure out how to determine the probability of failure or success of a circuit, made of multiple links. Like this:
    Node A-------link 1--------Node B---------link 2--------Node C
    What is the chance a message getting to Node C from Node A if the chance of an individual link failing is .1?
    Now, each link has an equal probability of failing.
    Only one link needs to fail for the entire circuit to fail. But because multiple links could fail, their probabilities aren't mutually exclusive, so I can't just add the probabilities of the individual failures.
    However since the failure of one link doesn't impact the probability that any other fails, they are independent meaning that the probability the probability of failure of link 1 and link 2 is P(f1 ∩ f2 ) = P(f1)P(f2)

    So if i try to determine the probability of success of the circuit, that means both links must not fail, which is equivalent to P(f1 ∪ f2)' = P(f1' ∩ f2')
    So the probability of success is:
    P(f1 ∪ f2)' = P(f1' ∩ f2')
    P(f1 ∪ f2)' = (1-P(f1))(1-P(f2))
    P(f1 ∪ f2)' = (1-.1)(1-.1) = .9 x .9
    P(f1 ∪ f2)' = . 81

    Which agrees with my Venn diagram!

    So I guess the key here was that the link failures are independent of each other and that's why I could use this technique.

    Thanks a lot for the help guys. Hopefully I'll understand this better with some more practice.
  11. Oct 11, 2016 #10
    Oops. Correction on the Venn diagram.

  12. Oct 11, 2016 #11


    User Avatar
    Science Advisor

    Seems correct.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted