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Probability question

  1. Mar 1, 2013 #1
    In a sample space Z = {a, b, c} is the probability distribution of the numbers x, y, 0.3 (in that order). What should be they that any {a, b} and {a, c} are independent?

    maybe i need a suggestion if the type i want to use down is correcty or not cuz i stucked...



    I know that for A1,A2,An we have this type : P(A1andA2and....andAn)=P(A1)P(A2).....P(An) but how i can use this if i have sets?
     
  2. jcsd
  3. Mar 1, 2013 #2

    Simon Bridge

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    You need a to be independent of b and c, but b and c need not be independent of each other?

    What is the definition of "independent" in this context? (Hint: in terms of probabilities.)
    You have:
    ... you don't have to apply the relation to whole sets, the A1,A2,... form a set {A1, A2, ...}.

    if A and B are independent, then P(A)+P(B)=?
     
  4. Mar 1, 2013 #3

    haruspex

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    If I'm interpreting this correctly, you need to find x and y such that P[f∈{a,b}|f∈{a,c}] = P[f∈{a,b}]. Is that it?
     
  5. Mar 1, 2013 #4

    Simon Bridge

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    Well if you are interpreting it correctly - then how would the relation P[f∈{a,b}|f∈{a,c}] = P[f∈{a,b}] possibly hold? After all, the two sets have a member in common.

    It boils down to how you read this bit:
    What should be they that any {a, b} and {a, c} are independent?
    ... which is "they" and what is "independent" of what?

    Presumably "they" is x and y.
    "{a,b} is independent", then I'd read that as outcome a is independent of outcome b... I don't think that {a,b} can be independent of {a,c} since they both contain a.

    I read the question as saying that {a,b,c} may or may not not be independent, but {a,b} and {a,c} are.
    Given that P(c)=0.3, what is P(a) and P(b)?

    However: you are closer to the course than I am so you may have an extra insight.
     
  6. Mar 1, 2013 #5

    haruspex

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    There is exactly one setting of x and y that makes it true.
    Aren't a, b and c atomic events? If so, they are therefore mutually exclusive, not independent. In probability space terms, a set of possible outcomes, like {a, b} constitutes a (nonatomic) event, and you want the events {a, b}, {a, c} to be independent.
     
  7. Mar 2, 2013 #6

    Simon Bridge

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    Because Z is a sample space? That would make sense OK.
    I still think the phrasing is sloppy... which is why I'm uncertain about how I was reading the question.
     
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