YLL討論網

chichali      發表於: 2003/6/11 下午 12:50:43              
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從一個裝有1~7號球的袋中依序取出n球，取後放回，試求點數和
為4n+1的機率為何？(以n來表示)  

ming          回覆於: 2003/6/11 下午 02:45:09                        
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根據驗證,應是6^(n-1)/7^n, 但不會証明  

ming          回覆於: 2003/6/11 下午 03:07:48                        
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sorry,錯了,之前只算了n=1,2,3
n=4,p(e)=191/7^4  

Meowth          回覆於: 2003/6/11 下午 04:18:01                        
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((x+xx+...+x^7)/7)^n 之(4n+1) 次方項係數
=((1+x+...+x^6)/7)^n 之(3n+1) 次方項係數

(1+x+...+x^6)^n
=(1-x^7)^n/(1-x)^n
=(1-C(n,1)x^7+C(n,2)x^14-....+C(n,n)(-x)^7n))/(1-x)^n

F(n)=(3n+1) 次方項係數
=H(n,3n+1)*1-H(n,3n-6)C(n,1)+H(n,3n-13)C(n,2)-...

P(n)=F(n)/7^n
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P(1)=H(1,4)*1 /7=1/7
P(2)=(H(2,7)*1-H(2,0)*2)/49=6/49
P(3)=(H(3,10)*1-H(3,3)*3)/343=36/343
P(4)=(H(4,13)-H(4,6)*4)/7^4=224/7^4
P(5)=1420/7^5
P(6)=9156/7^6
P(7)=59710/7^7


請以統計觀點,給出 P(n)的近似式!!

請以統計觀點,給出 P(n)的近似式!!
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mean=4
sd^2=((1-4)^2+(2-4)^2+...+(7-4)^2)/7=4
sd=2

deviation of mean=[(4n+1)-4n]/n=1/n
sd of mean=sd/sqrt(n)=2/sqrt(n)
x=1/n/(2/sqrt(n))=1/2sqrt(n)
xx/2=1/(8n)

P(n)~1/sd/sqrt(2*pi*n)/e^(xx/2)
=1/2/sqrt(2*pi*n)/e^(1/(8n))

Then do "Meowth 修正":
P(n)~ 1/2/sqrt(2*pi*(n+0.3))/e^(1/(8n))

when n>=3, error % < 0.4 %
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N, True value, 統計觀點 Estimation, Meowth修正
1, 0.142857143 0.176032663 0.154390859
2, 0.12244898 0.132501766 0.123558534
3, 0.104956268 0.110464782 0.105324037
4, 0.093294461 0.096667029 0.093233962
5, 0.084488606 0.087003697 0.084505461
6, 0.077824716 0.079754766 0.077832682
7, 0.072503804 0.07405865 0.072520933
8, 0.068145978 0.069430329 0.068164016
9, 0.064488436 0.065573286 0.064506983
10, 0.061362792 0.062294742 0.061380833