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Probability Axiom Question

  1. Apr 30, 2006 #1
    Hi

    I have this here probability axiom which I'm not sure what I have understood correctly.

    Let [tex]B_1 \ldots B_n[/tex] be independent events

    Then [tex]P(B_1 \mathrm{U} \ldots \mathrm{U} \ B_n) = 1[/tex] which is the same as

    [tex]P(B_1) + P(B_2) + \ldots + P(B_n) = 1[/tex]

    I would like to show that this only is valid if [tex]1 \leq k \leq n[/tex] such that

    P(B_k) = 1.

    Proof:

    If [tex]1 \leq k \leq n[/tex], then [tex]P(E) = 1 [/tex](where E is the probability space).

    Thereby it follows that [tex]P(B_1 U \ldots U \ B_n) = P(B_1) + P(B_2) + \ldots + P(B_n) = 1[/tex]

    This can be written as the [tex]\sum_{n=1} ^{k} P(B_{n+1}) = P(B_{k}) =1[/tex]

    Am I on the right track here?

    Best Regards
    Fred
     
    Last edited: Apr 30, 2006
  2. jcsd
  3. Apr 30, 2006 #2

    Hurkyl

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    I'm very confused by what you've written.


    So I will just state some facts:

    [tex]P(B_1 U \ldots U \ B_n) = P(B_1) + P(B_2) + \ldots + P(B_n)[/tex]

    is true when the [itex]B_i[/itex] are disjoint -- this equation is usually false when they are independent.


    If E is the event consisting of all possible outcomes, then P(E) = 1.

    In fact, P(A) cannot be greater than 1 for any event A.
     
  4. Apr 30, 2006 #3
    Hi Herkyl and thank You for Your answer,

    I have looked at it again and come to the conclusion that the proof should have said:

    Let B_1 \ldots B_n be independent events. Show that

    P(B_1 \mathrm{U} \ldots \mathrm{U} B_n)= 1, if and only if there exists a number 1 \leq k \leq n, such that P(B_k) = 1.


    Proof:

    If [tex]1 \leq k \leq n[/tex], then [tex]P(E) = 1 [/tex](where E is the probability space).

    Thereby it follows that [tex]P(B_1 U \ldots U \ B_n) = P(B_1) + P(B_2) + \ldots + P(B_n) = 1[/tex]
    Am I on the right path here now?
     
  5. Apr 30, 2006 #4

    AKG

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    Your theorem is false. "Heads" and "Tails" are independent events, neither P("Heads") nor P("Tails") is 1, but P("Heads" U "Tails") = 1
     
  6. Apr 30, 2006 #5

    Hurkyl

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    No, heads and tails are not independent. P(Heads and Tails) is certainly unequal to P(Heads)*P(Tails).
     
  7. Apr 30, 2006 #6
    Hello can I change my original theorem to make it true?

    If Yes, how?

    Sincerely Fred

     
  8. Apr 30, 2006 #7

    AKG

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    Okay, I guess I confused "independent" with "mutually exclusive".

    If you have two independent events B, B', then:

    [tex]P(B\cup B')[/tex]

    [tex] = P((B\cap B'^C) \sqcup (B\cap B') \sqcup (B' \cap B^C))[/tex]

    [tex] = P(B\cap B'^C) + P(B\cap B') + P(B' \cap B^C)[/tex]

    [tex] = P(B)P(B'^C) + P(B)P(B') + P(B')P(B^C)[/tex]

    [tex] = P(B)(1 - P(B')) + P(B)P(B') + P(B')(1 - P(B))[/tex]

    [tex] = 1 - [P(B) - 1][P(B') - 1][/tex]

    If [itex]P(B\cup B') = 1[/itex], then [P(B)-1][P(B') - 1] = 0, so either P(B) = 1 or P(B') = 1.
     
  9. Apr 30, 2006 #8

    HallsofIvy

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    If [itex]B_1, B_2, \ldots, B_n[/itex] are mutually exclusive then
    [itex]P(B_1 and B_2 and... and B_n)= P(B_1)+ P(B_2)+ \ldots + P(B_n)[/itex]
    follows from the definition of "mutually exclusive".
    If, in addition, they exhaust all mutually exclusive events, then
    [itex]P(B_1 and B_2 and... and B_n)= 1[/itex]

    ? How does that follow from the above? If P(Bk)= 1 then it follows that P(Bi)= 0 for any i not equal to k and so
    [itex]\sum_{n=1} ^{k} P(B_{n+1}) = P(B_{k}) =1[/itex][/quote] follows trivially. But you are claiming the converse: if the sum of probabilities is 1 then the probability for each i except one is 1- and that is not, in general, true.

     
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