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Confusing probability

  1. Nov 12, 2005 #1
    this is the problem...
    1. nine horses are entered in a horse race. If you "box" three horses (three are chosen and they can finish in any of the first 3 positions in the race), determine the probability that you will hold the winning ticket.
    -->i tried this one... (3C1*3C2*3C3)/9C3 but my answer is 3/28, which is wrong, it should be 1/14
    2. A drawer contains four red socks and five blue socks. Three socks are drawn one at a time and then put back before the next selection. Determine the probability that
    i. exactly red socks are selected
    --> i did this... [(4C2*5C3) + (4C1*5C4) + (4C0* 5C5)]/9C3 = 27/28, but it is wrong since the answer in the book is 304/729
    ii. at least two red socks are selected
    --> i dont get this one..
     
  2. jcsd
  3. Nov 12, 2005 #2

    Tide

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    There are 6 ways of holding the winning ticket and C(9,3) total ways of selecting any three horses for win, place and show.
     
  4. Nov 13, 2005 #3
    For the Horse problem the answer is 3P3/9C3
    (not sure if it is P cause in portugal we use A but in my calculator it says P it means the number of possible combinations of 3 horses in 3 different places where order counts (3!/(3-3)! = 3!)

    The second problem we can deal with it with binommial distribution:
    being x the number of consecutive red socks u take, u have
    n=3; p=4/9
    if u want to know the porbability of getting 1 red sock:
    p(x=1)=3C1*(4/9)^1*(5/9)^2=300/729

    for at least 2 red socks
    p(x>1)=p(x=2)+p(x=3)=3C2*(4/9)^2*(5/9)+3C3*(4/9)^3=304/729
     
  5. Nov 14, 2005 #4
    Sorry, my previous solution on the socks problem is wrong. I'm rly sry but i was really sleepy. Binomial distribution doesn't apply cause each act of choosing a sock is not indepandent from the others (Choosing 1 sock makes the probability of the next sock different). I also don't get the first question: do u want the probability of how many red socks?
     
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