Layout for defining prob. for Independent Events.

  • Thread starter Bacle
  • Start date
  • #1
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
mathman
Science Advisor
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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).
 
  • #3
662
1
""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).
 
  • #4
mathman
Science Advisor
7,761
415
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).
 
  • #5
662
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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.
 
  • #6
662
1
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.?
 
  • #7
mathman
Science Advisor
7,761
415
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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