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Probability independence problem

  1. Apr 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Let
    [itex]E= A \cup \bar{B} [/itex] and [itex]F= \bar{D} \cup C [/itex]
    Assuming that A,B,C,D are independent show that
    F and E are independent

    2. Relevant equations
    By definition A and by are independent if and only P(AB)=P(A)P(B).
    3. The attempt at a solution
    I tried to use set theory to simplify E[itex]\cap[/itex]F.But I couldn't imply the definition,all I got was this bunch of unions :
    EF=[itex] A \bar{D} \cup A C \cup \bar{B} \bar{D} \cup \bar{B} C[/itex]
     
  2. jcsd
  3. Apr 2, 2012 #2

    tiny-tim

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    hi naptor! :smile:

    start with, what is [itex]P(A \cup \bar{B})[/itex] ? :wink:
     
  4. Apr 2, 2012 #3
    Hi tim :smile thanks for the reply

    ok so :[itex]P(A \cup \bar{B})=P(A)+P(\bar{B})-P(A \cap \bar{B})[/itex] I can get rid off [itex]A \cap \bar{B}[/itex] using [itex]P(A)=P(A\cap B )+P(A \cap \bar{B}) \Rightarrow P(A \cap \bar{B}) =P(A)-P(A\cap B )[/itex] now I can plug this in my first eq and use the hypothesis I got:[itex]P(A \cup \bar{B})=P(A)+P(\bar{B})-P(A)P(\bar{B})[/itex] and the same thing for [itex]P(\bar{D}\cup C)[/itex]. I'm trying to combine these expressions in oder to get [itex]P(E \cap F)=P(E)P(F)[/itex].
    Am I on the right track?
     
    Last edited: Apr 2, 2012
  5. Apr 2, 2012 #4

    tiny-tim

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    hi naptor! :smile:
    a bit long-winded, and you still have a P(A), which should have cancelled :redface:

    easier would have been [itex]P(A \cup \bar{B})=P(\bar{B})+P(A \cap B)[/itex] :wink:

    try again :smile:
     
  6. Apr 2, 2012 #5

    Ray Vickson

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    It is a lot easier to first recognize some easily-proven preliminary properties: (i) two events [itex] U \text{ and } V[/itex] are independent if and only if [itex] U \text{ and } \bar{V}[/itex] are independent; and (ii) If [itex]U, V, W [/itex] are independent, then [itex] U \text{ and } V \cup W [/itex] are independent, as are [itex] U \text{ and } V \cap W [/itex].

    Because of these properties it does not matter whether we use [itex] B \text{ or } \bar{B}[/itex] and it does not matter whether we use [itex] C \text{ or } \bar{C}. [/itex] So, we can ask instead whether [itex] A \cup B \text{ and } \bar{C} \cup \bar{D} [/itex] are independent, or equivalently, whether [itex]A \cup B \text{ and } C \cap D [/itex] are independent. This last form is easier to work with. We have
    [tex] P\{ (A \cup B)\cap( C \cap D) \} = P\{ (A \cap C \cap D) \cup (B \cap C \cap \D) \}\\
    = P\{A \cap (C \cap D) \} + P\{B \cap (C \cap D) \} – P\{ (A \cap B)\cap(C \cap D)\} \\
    = P\{A\} P\{C \cap D\} + P\{B\} P\{C \cap D\} – P\{A \cap B \}P\{ C \cap D\}\\
    = P\{ A \cup B\} P\{ C \cap D\},[/tex]
    so [itex] A \cup B \text{ and } C \cap D[/itex] are independent.

    RGV
     
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