probability (6) 20點

2007-11-06 8:23 am

A most amazing event occurred during the second round of the 1989 US Open at Oak Hill in Pittsford , New York. Four golfers --Doug Weaver , Mark Wiebe , Jerry Pate , and Nick Price --- made holes in one on the sixth hole. According to the experts, the odds against a PGA golfer making a hole in one are 3708 to 1 --- that is , the probability of making a hole in one is 1/3709.

更新1:

Determine the probability to nine decimal places that at least four of the 155 golfers playing the second round would get a hole in one on the sixth hole by using

更新2:

a) the binomial distribution b) the Poisson approximation to the binomial. c) What assumptions are you making in obtaining your answers in parts (a) and (b) ? Do you think that those assumptions are reasinable? Explain your reasoning.

回答 (1)

魏王將張遼

2007-11-06 7:13 pm

✔ 最佳答案

(a) To find the probability that at least four get a hole in one by binomial distribution, we have:
P(At least 4) = 1 - P(None) - P(One) - P(Two) - P(Three)
Now,
P(None) = (3708/3709)155 = 0.959065530
P(One) = 155C1 × (1/3709) × (3708/3709)154 = 0.040090387
P(Two) = 155C2 × (1/3709)2 × (3708/3709)153 = 0.000832513
P(Three) = 155C3 × (1/3709)3 × (3708/3709)152 = 0.000011450
Hence,
P(At least 4) = 1 - 0.959065530 - 0.040090387 - 0.000832513 - 0.000011450
= 0.000000118
(b) If using Poisson distribution:
λ = 1/3709
Again, P(At least 4) = 1 - P(None) - P(One) - P(Two) - P(Three)
Now,
P(None) = λ0 × e-λ/0! = 0.999730421
P(One) = λ1 × e-λ/1! = 0.000269541
P(Two) = λ2 × e-λ/2! = 0.000000036
P(Three) = λ3 × e-λ/3! = 0.000000000003 (12 decimal places)
Hence,
P(At least 4) = 1 - 0.999730421 - 0.000269541 - 0.000000036 - 0.000000000003
= 1 (If corr. to 9 d.p.)
(c) The assumptions may include same physical conditions such as weather and also the match morale of those players. Both of these assumptions are impractical in reality since conditions are continuously varying.

參考: My Maths knowledge

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