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2D probability question

  1. Feb 2, 2010 #1
    we throw 3 coins

    y represents the throw of three coins
    x represents the throw of the first two
    we count the number of hetz gotten from the thrown coins.

    the solution says for the slot that:
    P(x=1,y=2)=0.25
    but i cant understand how
    ?
    the logical solution says:
    for x to be 1 we have 01 10
    so for y to be 2 011 101
    so with respect to y we have two possibilities from 8

    but the formal mathematical says

    P(x=1,y=2)=P(x=1)*P(y=2/x=1)
    for P(x=1) we need to have 01 10 which is 1/2 or 0.5
    for P(y=2/x=1) its 2/8 or 0.25

    P(x=1,y=2)=P(x=1)*P(y=2/x=1)=0.5*0.25=0.125
    and not 0.25
     
  2. jcsd
  3. Feb 2, 2010 #2
    looks like the only mistake is in this last line. p(y=2|x=1) would be 1/2

    the "logical" solution is fine, and the best way to do the problem.
     
  4. Feb 2, 2010 #3
    why p(y=2|x=1) 1/2 ?
    3 digit gives 8 possibilities
    our possibilities are 011 101
    so its 1/4
     
  5. Feb 2, 2010 #4

    vela

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    You're mixing up p(y=2 and x=1) and p(y=2|x=1). The probability p(y=2 and x=1) is 2/8, but p(y=2|x=1) = p(y=2 and x=1)/p(x=1) = (2/8)/(1/2) = 1/2.

    Intuitively, when you're given x=1, the possible outcomes are restricted to 010, 011, 100, and 101, so you divide by 4 instead of 8.
     
  6. Feb 3, 2010 #5
    thanks i understand now
     
    Last edited: Feb 3, 2010
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