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Odds help please

  1. Jul 29, 2008 #1
    With three 50/50 odds choices, what are the odds of getting any 2 correct?
     
  2. jcsd
  3. Jul 29, 2008 #2

    berkeman

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    Staff: Mentor

    What are your thoughts? It also depends on whether each choice is independent of the previous choices (like coin flips are). This is a bit too much like homework/coursework, so I'm moving it to Homework Help.
     
  4. Jul 29, 2008 #3
    I'm far too old for it to be homework. Each choice is independant. If a person was guessing the colour of randomly presented unseen playing cards, what are the chaces of guessing any two of the thee correctly. I think it must be less than 1 in 4.
     
    Last edited: Jul 29, 2008
  5. Jul 29, 2008 #4

    berkeman

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    Staff: Mentor

    Hey, I'm pretty old as well, and do lots of homework! :biggrin:

    The key is to think of it as a decision tree. You can write it out like this, with a correct pick = 1, and a wrong pick = 0:

    Code (Text):

    1st 2nd 3rd Total
     0    0    0     0
     0    0    1     1
     0    1    0     1
     0    1    1     2
     1    0    0     1
     1    0    1     2
     1    1    0     2
     1    1    1     3
     
     
    There are 8 possible outcomes. How many of them result in getting 2 right?

    Can you see how you would extend this to, say, a 60/40 chance of picking correctly?
     
    Last edited: Jul 29, 2008
  6. Jul 29, 2008 #5
    I have it - Thanks
     
  7. Jul 30, 2008 #6
    If you've learned the chose formula, then a good way to do problems that have specifically 50/50 odds is this:
    [tex]\frac{n C r}{2^n}[/tex] where n is the number of trials, and r is the number of successes. So, to answer your question, 3 C 2=3, and 23=8. So your answer should be [tex]\frac{3}{8}[/tex]
     
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