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Union of probabilities(coin tosses)

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Toss a coin 3 times. What is the probability that we get a head on the first toss or a head on the second toss or a head on the third toss?


    2. Relevant equations
    Pr(AorB)=Pr(A)+Pr(B)-Pr(AandB)



    3. The attempt at a solution
    A=head on 1st toss
    B=head on 2nd toss
    C=head on 3rd toss

    Pr(AorBorC)=Pr(A)+Pr(B)+Pr(C)-Pr(A&B&C)

    =(1/2)+(1/2)+(1/2)-(1/8)=11/8

    but 11/8>1 which is a contradiction. So something is wrong with my sol.
     
  2. jcsd
  3. Sep 26, 2009 #2

    Office_Shredder

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    This formula is true



    This one isn't. The formula above only works as written when you have precisely two sets. You can expand it to a more general setting by considering what it means though. P(A or B) = P(A) + P(B) - P(A and B)

    What this is really doing is just a counting argument. The number of things in the two sets A and B is just the number of things in A plus the number of things in B minus the number of things in both A and B. We had to subtract the last part because everything that was in A and B was counted when we counted the number of things in A, and also counted when we counted the number of things in B.

    So similarly, to find P(A or B or C), we need the number of things in A, plus the number of things in B, plus the number of things in C. But in this case we double counted everything that is in A and B, and we double counted everything in B and C, and we double counted everything in A and C for the same reason as above. Also, anything in all three of A and B and C was counted three times, but when we remove everything in A and B, and everything in A and C, and everything in B and C, we uncounted those things three times also. So we need to add back in the number of things in all of A, B and C at the same time.

    Do you see how to transform that into a formula about probabilities?
     
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