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Odds in Probability of Green Eyes

  1. May 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Of 27 students in a class, 11 have blue eyes, 13 have brown eyes, 3 have green eyes. If 3 students are chosen at random what are the odds of the following occurring?

    Only 1 has green eyes.


    2. Relevant equations
    Odds = P( s )/P( f ) = P( s )/( 1 - P( s ))
    Probability success = P( s )
    Probability fail = P( f )



    3. The attempt at a solution

    P( s ) = (3/27) * (24/26) * (23/25) = 92/975
    P( f ) = 1 - P( s ) = 883/975
    Therefore: Odds = 92/883
    I went over this problems many many times, and i believe this is the correct answer, but the back of the book says the answer is 92/233. There is a chance the back of the book is wrong. any ideas? thanks a lot.
     
  2. jcsd
  3. May 23, 2009 #2
    Nevermind. there are 3 ways to get 1 green person (1st,2nd, or 3rd) which means P( s ) = 3*92/975. and when you work out P( f ) you find the odds are: 92/233. QED. thanks
     
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