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Simple Probability Question(HELP)

  1. Nov 26, 2007 #1
    1. The problem statement, all variables and given/known data

    Given that [tex]P(T) = P(U) = 0[/tex] then show that [tex]P(T \cup U) = 0[/tex]

    Given that [tex]P(T) = P(U) = 1[/tex] then show that [tex]P(T \cap U) = 1[/tex]


    2. Relevant equations

    I am told that I need to use the following equations.

    (1) [tex]P(T \cup U) = P(T) + P(U)[/tex] if [tex]P(T \cap U) = 0[/tex]

    (2) [tex]P(T \cup U) = P(T) + P(U) - P(T \cup U)[/tex]

    3. The attempt at a solution

    My Solution for question one.

    Since We know that [tex]P(T) = P(U) = 0[/tex] then by using eqution

    We get that [tex]P(T \cup U) = P(T) + P(U) = 0 + 0 = 0[/tex]

    My Solution for question two.

    Here I have a feeling that I need to use equation two, but how do I deduce
    [tex]P(T \cup U)[/tex] ??

    Best Regards
    Beowulf..
     
  2. jcsd
  3. Nov 26, 2007 #2
    The last term should be [itex]P(T \cap U)[/itex].

    How do you know to use equation (2)? What if [itex]P(T \cap U) \ne 0[/itex]?

    Why do you have a feeling? Describe your reasoning.
     
  4. Nov 26, 2007 #3
    From what I see it [itex]P(T \cap U) [/itex] cannot be larger than one for obvious reasons. Because that would give a negative probability.

    I am not sure about that, another way of proving question (2)

    Could it say If P(C) is the entire probability space where then since P(T) = P(U), then I could say [tex] P(C) = P(U) + P(T) = 2 [/tex] and then [itex]P(T \cap U) =P(U) \cdot P(T) = 1[/itex]
    Then [itex]P(T \cup U) = P(c) - P(U) \cdot P(T) = 1[/itex]

    Is my argument my reasonable? If yes can it be used in (1) too?

    Best Regards

    Beowulf
     
    Last edited: Nov 26, 2007
  5. Nov 26, 2007 #4
    I will write [itex]T \cap U[/itex] as TU for short. If I understand you correctly, you're saying that P(TU) cannot be greater than 1 because if it is, it's negative?!

    By the axioms of probability, P(C) = 1, not 2. Note that P(U) and P(T) have the same probability. What does that say about C, U and T? Also P(TU) = P(T)P(U) if T and U are independent events. The problem does not mention that T and U are independent.
     
  6. Nov 26, 2007 #5
    Hello Again,

    The whole problem is as follows: Let (S, E, P) be a probability space and let T and U be events(Nothing is said about the events).

    Show, that if

    (1) [tex]P(T) = P(U) = 0[/tex] then [tex]P(T \cup U) = 0[/tex]

    (2) [tex]P(T) = P(U) = 1[/tex] then [tex]P(T \cap U) = 1[/tex]

    My Solultion(1):[tex]P(T \cup U) = P(T) + P(U) = 0 + 0 = 0[/tex]

    My Solultion(2):[tex]P(T \cap U) = P(T) \cdot P(U) = 1 \cdot 1 = 1[/tex]

    Could this be it? What else is there to add?

    Sincerely Yours
    Beowulf
     
  7. Nov 26, 2007 #6
    This will only work if P(TU) = 0. Can you prove that?

    This will only work if T and U are independent. Can you prove that?
     
  8. Nov 26, 2007 #7
    Question(a)

    This will only work if P(TU) = 0? By this you mean proving that P(TU) = emptyset.

    The only that I know here is the that is the consequence of the formula P(T U U) = P(T) + P(U).

    Is that what you mean??

    Question(b)

    Since nothing is said about the events being indepedent do I assume they are??

    Finally if these two can be proved, does that then complete the solution for (1) and (2)??

    Many thanks in advance.

    Sincerely Beowulf.
     
    Last edited: Nov 26, 2007
  9. Nov 26, 2007 #8
    TU is not necessarily empty. Consider the elements in TU individually and their probabilities when considered as events.

    No. Forget about independence for the moment. If P(X) = 1 where X is some arbitrary subset of the sample space, what can you say about X?

    All you have to do is show that your results follow from the axioms of probability.
     
  10. Nov 27, 2007 #9
    Regarding (1) do I considered the events to the mutually exclusive?? And therefor if P(T) occurs then P(U) also occurs and thus P(T u U ) = P(T) + P(U)?

    Or is the explaination more simple??

    In (2) Is it something to do with the events being mutually inclusive??

    SR

    Beowolf
     
  11. Nov 27, 2007 #10
    You're going about it the wrong way. For problem (1) you're told that P(T) = P(U) = 0. What does this say about T and U? Ditto for problem (2).
     
  12. Nov 27, 2007 #11
    Just so we understand each other. Do You mean?

    [tex]T \subseteq U[/tex] and [tex]U \subseteq T[/tex]

    Therefore their union is defined as

    [tex]T \cup U = T + U[/tex]

    Therefore their intersection is defined as

    [tex]T \cap U = T \cdot U[/tex]

    And then use this fact to claim that my result in (1) and (2) are true!

    How does that sound?

    BeoWulf
     
    Last edited: Nov 27, 2007
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