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How do I expand this probability union thing

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data

    I need to expand this:
    P(A∪B∪C∪D)
    A,B,C,D are not disjoint.


    2. Relevant equations



    3. The attempt at a solution

    P(A∪B∪C∪D) = P(A) + P(B) + P(C) + P(D) - P(A∩B) - P(A∩C) - P(B∩D) - P(C∩D) + P(A∩B∩C∩D)

    Is that right
     
  2. jcsd
  3. Oct 4, 2011 #2

    LCKurtz

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    Not quite. You are missing some terms and it isn't finished. You have to subtract terms for all pairs. For example you are missing -P(A∩D). Then you have to add all the triple intersections then subtract that last term. Here's the formula near the bottom of the page:

    http://mathworld.wolfram.com/Probability.html
     
  4. Oct 4, 2011 #3
    I don't get it. Those weird symbols are confusing.
    So is it

    P(A∪B∪C∪D) = P(A) + P(B) + P(C) + P(D) - P(A∩B) - P(A∩C) - P(A∩D) - P(B∩C) - P(B∩D) - P(C∩D) + P(A∩B∩C) + P(A∩B∩D) + P(B∩C∩D) + P(A∩B∩C∩D) ?
     
  5. Oct 4, 2011 #4

    LCKurtz

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    Yes except you missed P(A∩C∩D) and the last term needs a minus sign.
     
  6. Oct 5, 2011 #5
    So I minus the last term because I'm adding 4 things?
    If it's 3 things I add?
     
  7. Oct 5, 2011 #6

    LCKurtz

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    Yes. The signs alternate in that fashion.
     
  8. Oct 5, 2011 #7
    why?
     
  9. Oct 5, 2011 #8

    LCKurtz

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    For the same reason you get a minus in the expansion

    P(A∪B) = P(A) + P(B) - P(A∩B)

    A and B both contain A∩B so the first two terms account for A∩B twice. So you have to subtract it out to avoid counting it twice. You get a similar thing with three and four; it's just a little more complicated to keep track of.
     
  10. Oct 5, 2011 #9
    wow that's so annoying.
    so if I had like 20 terms it's hard to do by hand
     
  11. Oct 5, 2011 #10

    LCKurtz

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    Try not to lose any sleep over it. :cry:
     
  12. Oct 6, 2011 #11
    Hey zeion - you must be taking the same stats course as me. Make sure to notice that the question specifies that the events are independent!
     
  13. Oct 6, 2011 #12
    Okay so does that mean I have to do something else
    I'm so confused
     
  14. Oct 6, 2011 #13
    Since they're independent,

    P(A∪B∪C∪D) = P(A) + P(B) + P(C) + P(D)

    Also,

    P(A∩B∩C) = P(A)P(B)P(C) and

    P(A|B) = P(A)

    I find that the text he quotes from (there's a link to the PDF version of it on the course site) is very useful. Not that I'm having an easy time of it!
     
  15. Oct 6, 2011 #14

    LCKurtz

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    No no no! That's true for disjoint sets, which these are given not to be. They aren't given as independent and that isn't the same as disjoint anyway.
     
  16. Oct 6, 2011 #15
    Hi LCKurtz,

    I'm a newbie, so I may very well have it wrong, but the full question states:

    A school has a computer lab that is closed for the summer, when no students are
    there, and then is re-opened in the fall. Sometimes, a computer in the lab that worked before the lab closed is found to not work properly when it is powered up again in the fall.
    These computer failures are always due to one or more of the following causes:
    A) Failure of the power supply.
    B) Failure of the disk drive.
    C) Failure of the RAM.
    D) Failure of the CPU.
    From long experience, it is known that these failure events are INDEPENDENT, and that their probabilities of occurrence (for a single computer) are P(A) = 0.04, P(B) = 0.03, P(C) = 0.02, and
    P(D) = 0.01.
    1) Find the probability that a computer will fail for any of these reasons — that is, find P(F), where F = A ∪ B ∪ C ∪ D.
    2) Find the probability that a non-working computer has failed for more than one reason — that is, find the conditional probability that more than one of A, B, C, and D has occurred, given that F = A ∪ B ∪ C ∪ D has occurred.

    I'll have to dig through my notes to see where he told us that P(A∪B∪C∪D) = P(A) + P(B) + P(C) + P(D). Perhaps I misinterpreted.
     
  17. Oct 6, 2011 #16
    Independent is not disjoint. I believe the prof emphasized that point as well.
    Independent just means their likelihood of happening does not depend on each other, but they are not disjoint because more than one of them could happen at the same time.
    ie. "one or more of the following causes"

    Although I'm still not sure how the thing about them being mutually independent changes anything..
     
  18. Oct 6, 2011 #17
    Well, I think that means that these are still true:

    P(A∩B∩C) = P(A)P(B)P(C) and

    P(A|B) = P(A)
     
  19. Oct 6, 2011 #18
    My faulty memory - I had mixed up disjoint and independent. For independent, it seems like the easiest calculation is:

    P(A∪B∪C∪D) = 1 - [1-P(A)][1-P(B)][1-P(C)][1-P(D)]

    But don't quote me on that!!! Like I said - I'm a newb.
     
  20. Oct 6, 2011 #19
    Where did you get that from..?

    I hate this course.. pretty sure I failed that quiz.
     
  21. Oct 6, 2011 #20
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