f******g
发帖数: 423 | 1 据说是Bernoulli distribution，但我还是不知道如何套公式
已知有一组数共39个，value 为0 或1。
randomly select 20个，都是0。
问，1。所有数都是0的几率？
2。 如果要达到95% confidence level，我要抽查几个数？ |
g********5
发帖数: 62 | 2 You should ask this question in Statistics board.
I do not understand your second question. Solution to 1th question is:
Prob(all 39# are zero)
= (39 choose 20) / { (20 choose 20) + (21 choose 20) + ...+ (39 choose 20) }
= (39 choose 20) / (40 choose 21)
= 21/40
where (n choose k)= n!/{ k!(n-k)! } |
f******g
发帖数: 423 | 3 谢谢，我其实也不知道，是乱问的。没想到能算出具体的概率而不是一个范围，所以才
问了第2个问题。
我这就转过去。
}
【在 g********5 的大作中提到】
: You should ask this question in Statistics board.
: I do not understand your second question. Solution to 1th question is:
: Prob(all 39# are zero)
: = (39 choose 20) / { (20 choose 20) + (21 choose 20) + ...+ (39 choose 20) }
: = (39 choose 20) / (40 choose 21)
: = 21/40
: where (n choose k)= n!/{ k!(n-k)! }
|
w******i
发帖数: 503 | 4 your solution assumes the probabilities of 20 zero's, 21 zero's ... 39 zero'
s are same. should they be different if every number is equally likely to be
0 and 1? then prob(39 0's)/prob(20 0's) = (39 choose 39)/(39 choose 20)?
}
【在 g********5 的大作中提到】
: You should ask this question in Statistics board.
: I do not understand your second question. Solution to 1th question is:
: Prob(all 39# are zero)
: = (39 choose 20) / { (20 choose 20) + (21 choose 20) + ...+ (39 choose 20) }
: = (39 choose 20) / (40 choose 21)
: = 21/40
: where (n choose k)= n!/{ k!(n-k)! }
|
