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 【重奖征答】延长时间，提高奖金，透露最后答案。请大家继续尝试解题！！！ |
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【重奖征答】延长时间，提高奖金，透露最后答案。请大家继续尝试解题！！！ -- 偶而冒个泡 - (965 Byte) 2012-1-10 周二, 12:48 (4037 reads) |
偶而冒个泡
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作者：偶而冒个泡 在 海归商务 发贴, 来自【海归网】 http://www.haiguinet.com
Problem 1
Let B be the event that this particular sequence is observed in the 60 coin flips with 33 heads shown, Ca, Cb, Cc be the events that coin a,b,c is chosen from the box, respectively.
For any sequence of 33 heads and 27 tails,
P(B|Ca) = (1/3)^33*(2/3)^27
P(B|Cb) = (1/2)^33*(1/2)^27
P(B|Cc) =(2/3)^33*(1/3)^27
And we also know P(Ca) = P(Cb) =P(Cc) = 1/3
Therefore,
P(Cc|B) = P(Cc)*P(B|Cc)/[ P(Ca)*P(B|Ca) + P(Cb)*P(B|Cb) + P(Cc)*P(B|Cc)] = 0.002950
Likewise
P(Cb|B) = 0.8008229
P(Cc|B) = 0.188820
Remark: Above formula involving conditional probability is often called Bayes Rule.
Problem 2
The probability that the RED fish are the first to go extinct from the pond is the same as the probability that always BLUE or GREEN fish is caught while all the RED fish are still in the pond
Let L be the event that all the RED fish are still in the pond
P(B&G before R) = P(B)*P(L|B) + P(G)*P(L|G)+P(R)*P(L|R)=(53/204)*64/(64+87)+(64/204)*53/(53+87)+0 = 0.229
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You may also try exhaustion method to solve the problem:
Total # of the ways to catch all fish = (87+53+64)!/87!*53!*64!
you also need to figure out total # of the ways that the red fish are the first to go extinct from the pond, then divide by above number. The answer should be the same as 0.229.
作者：偶而冒个泡 在 海归商务 发贴, 来自【海归网】 http://www.haiguinet.com
上一次由偶而冒个泡于2012-1-13 周五, 08:27修改，总共修改了3次
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 【重奖征答】延长时间，提高奖金，透露最后答案。请大家继续尝试解题！！！ -- 偶而冒个泡 - (965 Byte) 2012-1-10 周二, 12:48 (4037 reads) - 【【我的解法（见内）。解法可以有多种，我的只是其中之一。大家可以评估一下自己的答案，觉得答对了，请自报出身来领取奖金】】 -- 偶而冒个泡 - (1396 Byte) 2012-1-12 周四, 14:11 (745 reads)
- 删。 -- 咳嗽 - (0 Byte) 2012-1-12 周四, 14:53 (444 reads)
- 第2题错了。 -- 绽铃子 - (27 Byte) 2012-1-13 周五, 00:02 (428 reads)
- 我觉得好难好难 -- 阳光在多城 - (118 Byte) 2012-1-11 周三, 22:27 (461 reads)
- 吼吼 -- nnzqj - (201 Byte) 2012-1-10 周二, 23:14 (515 reads)
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