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Calculating events from phrasal expressions

  1. Feb 25, 2017 #1
    1. The problem statement, all variables and given/known data
    This excresice is supposed to help you understand the basic operations of sets, later used in probability. I am given the following phrases and have to write them in using mathematics.
    Given three events A, B and C, which belong to sample space S, calculate the following events:
    a) Only A happens
    b) A and B happen, but not C
    c) At least one of the events happen
    d) At least two of the events happen
    e) All the events happen
    f) None of the events happen
    g) One of the events happen at most
    h) Two of the events happen at most
    i) Exactly two of the events happen
    j) Three of the events happen at most


    2. Relevant equations
    No needed.

    3. The attempt at a solution
    Here are my solutions. Of course I want to know if I'm correct or not, but I basically need some help with g, h and k. I can't quite understand these expressions and when I try to turn them into mathematical expressions I end up with nonsense. I also used Venn diagrams to understand each event better.

    a) A∩B'∩C'
    b) This translates to "Only A and B happen", so A∩B∩C'
    c) This translates to "A or B or C or all of them", so A∪B∪C
    d) "Two or more events happen at the same time", so (A∩B)∪(A∩C)∪(B∩C)
    e) A∩B∩C
    f) Meaning "... at the same time", so A'∩B'∩C' or (A∪B∪C)' from De Morgan's Law.
    g) I can't wrap my head around this one. I'd think it this way: This translates to "Two or more events can't happen at the same time", so I'd go full boolean and think " NOT(A AND B) AND NOT(A AND C) AND NOT(B AND C) AND NOT(A AND B AND C)". Is this correct? Is there any other way to make me understand it? The Venn Diagram helps me get it better, but I could still use some further explaination.
    h) This one seems easier to understand if I translate it to "Three events can't happen at the same time, so it's simply (A∩B∩C)' or (A'∪B'∪C') from De Morgan's Law. The Venn Diagram of this also makes sense to me.
    bfe3e02f2b074b4f9994c5063ec55fe3.png
    i) This one looks easy using the Boolean logic again. (A AND B AND NOT C) OR (A AND NOT B AND C) OR (NOT A AND B AND C).
    j) This makes the less sense, since there are three events. I honestly can't think of anything.
     
  2. jcsd
  3. Feb 25, 2017 #2

    haruspex

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    For g, compare it to d.
    For j, what notation have you been taught for the universal set? Maybe just "true"?

    The rest look fine.
     
  4. Feb 27, 2017 #3
    Hmmm, so g would look like h, only without the striped parts, right? And my "boolean" expression should be correct if we replace AND with ∩, OR with ∪ and NOT with '.
    a916c82d10c741759859517b8f8dc759.png
    About j, now. We usually refer to the universal set as S. Is this the answer? Because "3 events at most" means 3 or less and the events are exactly 3? In that case the expression should be (A∪B∪C)∪(A'∩B'∩C').
     
  5. Feb 27, 2017 #4

    haruspex

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    I still think it more fruitful to compare g with d.
    Are there any events that are ruled out by j?
     
  6. Feb 27, 2017 #5
    You mean because they're each other's complement? I can't see any other resemblence.
    Looks like not. "Three events at most" translates to "Three events or less". Since there are three events in the given sample space, more than three events can't happen even if we take the whole sample space.
    I'm not sure if you question me to make sure I understand or because I'm wrong.:biggrin:
     
  7. Feb 27, 2017 #6

    haruspex

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    Yes. You already solved d, so that gives you an easy way to write an answer for g.
    You did not seem to be sure that the answer is S, so I was just trying to get you to think in slightly different terms. Or maybe you are uncomfortable expressing the answer as S?
     
  8. Feb 28, 2017 #7
    I kind of am uncomfortable, because I needed to understand this in A, B and C terms. Which is weird if you try to express it like this from scratch, but once you realise it's S you can easily jump to the expression.
    Thanks a lot for your help and time!
     
  9. Feb 28, 2017 #8

    haruspex

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    I suppose an alternative would be A∪A'.
     
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