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Layout for defining prob. for Independent Events.

  1. Mar 31, 2010 #1
    Hi, everyone:

    I was reviewing some intro material, and I ended up confused with the issue of

    independence with the following problem:

    We are given two teams, A,B, playing against each other. A wins with probability

    P(A)=0.6 , B wins with P(B)=0.4 (games are played until someone wins.). We also

    assume that the outcome of any game is independent from that of any other

    game.


    **Now** . Here is where I am confused:

    Let a string with A in i-th place and B in j-th place denote that A won the i-th

    game and B won the j-th game.

    I am trying to show that the probability of team A winning two consecutive games

    is (0.6)^2 , arguing that the outcome : AA has probability (0.6)(0.6) , since the

    outcome of game 2 is (assumed) independent from that of game 1.

    **BUT** I am having trouble expressing the event 'AA' as an intersection of two

    events E,E' , which I need to do in order to use the rule: P(E/\E')=P(E)*P(E') , (with

    /\:= intersection , and * is product)

    My sample space is :

    { A,B , AA, AB, BA, BB, AAA, ....}


    And I don't see how to express 'AA' as an intersection of events in order to justify

    saying that the probability of 'AA' is (0.6)(0.6) . Any Ideas.?.

    Thanks.
     
  2. jcsd
  3. Mar 31, 2010 #2

    mathman

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    You are confusing the teams A and B with the events A wins or B wins. The outcome of the first game may be called O1 and the second game O2. So the independent events of interest are Oi=A wins for i=1 and 2.
    So P(O1=A wins and O2=A wins)=P(O1=A wins)P(O2=A wins).
     
  4. Apr 2, 2010 #3
    ""You are confusing the teams A and B with the events A wins or B wins."


    Not really. I described my choice of notation:


    " Let a string with A in i-th place and B in j-th place denote that A won the i-th

    game and B won the j-th game."


    By this notation, a string AA denotes the team A winning twice consecutively.


    But the issue I had was that my options for deriving a probability were:

    0) Define the sample space S. Then an event E is an element in 2^S.

    i) P(E union E')=P(E)+P(E')

    ii) P (E intersection E') =P(E)*p(E')

    And I do not see any way of expressing 'AA' as neither of i nor ii).
     
  5. Apr 2, 2010 #4

    mathman

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    Your notation is what is making it difficult for yourself. Let A1 be the event A won the first game and A2 be the event A won the second game. So P(A1 and A2) = P(A1)P(A2).
     
  6. Apr 2, 2010 #5
    No. The point is I can only use strict set intersection, or strict set union,

    and not the informal ( but sometimes equivalent) 'E or E' , or E intersection E'.

    Can you express the event of A winning twice as a _set intersection_ of

    events.?.

    I mean, define a sample space S of all outcomes, then take 2^S to be the

    set of events.

    Can you then express A winning twice as an intersection of any two elements in 2^S.?.


    I don't mean in the form 'A and B' , but as an actual intersection of elements of 2^S.
     
  7. Apr 2, 2010 #6
    Just to clarify my approach:

    My sample space would be :

    {A,B ; AA, AB, BA, BB ; AAA, AAB, ABA, ABB, BBB, ,BAA, BAB, BBA........}

    And I don't see how 'AA' itself is an intersection. I am trying to avoid

    product spaces.


    I can see how to use intersection when, e.g., I throw a single fair

    die, and I get the sample space : {1,2,3,4,5,6} , then 2^S is my

    sample space, and the event : "even and larger than 3" is the intersection

    of , e.g., { 2,4,6} and {4,6}. But once I throw the die twice , expressing

    outcomes as intersections becomes more difficult. Can you express

    an event as an intersection in this case.?
     
  8. Apr 3, 2010 #7

    mathman

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    Science Advisor

    Each trial is a possible event, so a particular result of two trials is the intersection of the outcomes of each trial.
     
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