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Math Help - Golf & Probability Puzzle

  1. #1
    TurkeyRun
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    Golf & Probability Puzzle

    Can someone help me with the following question:

    In the 1989 US Open Golf Championship, 4 golfers made hole-in-one on the 6th hole.The probability of a professional golfer making a hole in one (on any given hole) is estimated to be 1/3709.

    There were 155 golfers participating in the second round. Assuming the estimate given above, what is the probability that exactly 4 of the 155 golfers would get a hole-in-one on the sixth hole?
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  2. #2
    Newbie
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    Jul 2005
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    Assuming each hole is independent from the next and that the probability of getting a hole in one remains constant from hole to hole, this question is simply a case of using the Binomial Distribution.

    So the probability that exactly 4 of the 155 golfers got a hole-in-one on hole 6 is:

    p(x)=(n/x) p^x (1-p)^n-x

    where

    (n/x)= n!/x!(n-x)!

    so p(4)=0.0000001

    It was basically a fluke!
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  3. #3
    Super Member Rebesques's Avatar
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    e?

    the probability of getting a hole in one remains constant from hole to hole
    That never happens. If there's no luck at start, there will be none later. ..
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  4. #4
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    I think this is just a sure luck for these 4 golfer. I mean they have to be like math geniuses to calculate their probability on that golf course. I really thinks this is just a mere coincidence.

    I love <a href="http://buydiscountgolfballs.com">Discount Golf Balls</a>
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