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| | 由于X <math>\sim B(n, p)</math>和Y <math>\sim B(X, q)</math>，由<font color="#ff8000">全概率公式 the law of total probability </font>, | | 由于X <math>\sim B(n, p)</math>和Y <math>\sim B(X, q)</math>，由<font color="#ff8000">全概率公式 the law of total probability </font>, |
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| | <math> \widehat{p\,} + \frac{z^2}{2n} + z</math> | | <math> \widehat{p\,} + \frac{z^2}{2n} + z</math> |
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| − | <math> &= \sum_{k=m}^n \binom{n}{k} \binom{k}{m} p^k q^m (1-p)^{n-k} (1-q)^{k-m}</math>
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| | <math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math> | | <math>\frac{\widehat{p\,}(1 - \widehat{p\,})}{n} </math> |
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| | <math> 1 + \frac{z^2}{n}</math> | | <math> 1 + \frac{z^2}{n}</math> |
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| − | <math>\Pr[Y = m] &= \binom{n}{m} p^m q^m \left( \sum_{k=m}^n \binom{n-m}{k-m} p^{k-m} (1-p)^{n-k} (1-q)^{k-m} \right) \\[2pt]}</math><ref><nowiki>{{cite book</nowiki>
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| − | <math> &= \binom{n}{m} (pq)^m \left( \sum_{k=m}^n \binom{n-m}{k-m} \left(p(1-q)\right)^{k-m} (1-p)^{n-k} \right)</math>
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| − | | chapter =Confidence intervals
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| − | | chapter-url =http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm
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| − | After substituting i = k - m in the expression above, we get
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| − | | title =Engineering Statistics Handbook
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| − | \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right)
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| − | | publisher =NIST/Sematech
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| − | Notice that the sum (in the parentheses) above equals (p - pq + 1 - p)^{n-m} by the binomial theorem. Substituting this in finally yields
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| − | | year =2012
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| − | 1.1.1.2.2.2.2.2.2.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.3.4.3
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| − | | access-date =2017-07-23
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| − | <nowiki>}}</nowiki></ref>
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| | ====比较==== | | ====比较==== |