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Probability homework problem help

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data

    a singlecard is drawn from a pack of 52 . what is the probability P((oddUblack)Ueven)

    2. Relevant equations



    3. The attempt at a solution

    P(oddUblack) = 26/52 + 26/52 - 13/52 = 39/52

    so is the next part mutually exclusive( if you draw and odd in the first part it cant be even?)

    so my probability is 39/52 + 26/52 which obviously isnt right...
     
  2. jcsd
  3. Mar 14, 2012 #2

    tiny-tim

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    hi gtfitzpatrick! :smile:
    but P((oddUblack)Ueven) = P(oddUeven) = 100% :confused:
     
  4. Mar 14, 2012 #3
    Re: probability

    but then Uwith black so does that mean 0.5?
     
  5. Mar 14, 2012 #4
    Re: probability

    are the laws distributive can i say p((oUe)Ub)?
     
  6. Mar 14, 2012 #5

    tiny-tim

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    are you sure you have the question right?

    it asks for odd or black or even …

    everything is odd or black or even​
     
  7. Mar 14, 2012 #6
    Re: probability

    thanks for help Tim.
    The question is definatly P((ODD U BLACK) U EVEN)
    I can see the way your thinking but do you think the brackets change it?
     
  8. Mar 14, 2012 #7

    HallsofIvy

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    Re: probability

    Is it possible that at least one of your "[itex]\cup[/itex]" is supposed to be "[itex]\cap[/itex]"?
     
  9. Mar 14, 2012 #8
    Re: probability

    no they are both U, do you think its a mistake?
     
  10. Mar 14, 2012 #9

    tiny-tim

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    [itex]Y\cap p![/itex] :biggrin:
     
  11. Mar 14, 2012 #10
    Re: probability

    lol very funny :)
     
  12. Mar 14, 2012 #11

    HallsofIvy

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    Re: probability

    If they are all [itex]\cup[/itex] then the problem is trivial as tiny-tim has told you. [itex]E\cup U[/itex] includes all cards so "adding" all black cards doesn't change anything.
     
  13. Mar 14, 2012 #12
    Re: probability

    cool, thanks chaps
     
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