math Question
2007-10-10 7:35 pm
show for the binomial distribution that the probability of x successes in n trials, when the probability of one success is p is the same as the probability of (n-x) successes in n trials when the probability of oned success is (1-p).
回答 (1)
2007-10-10 8:05 pm
✔ 最佳答案
the probability of x successes in n trials, when the probability of one success is p= (nCx) (p^x) [(1-p)^(n-x)]
the probability of (n-x) successes in n trials when the probability of oned success is (1-p)
= [nC(n-x)] [(1-p)^(n-x)] {p^[n-(n-x)]}
= (nCx) [(1-p)^(n-x)] p^[x)
=(nCx) (p^x) [(1-p)^(n-x)]
Right?
收錄日期: 2021-04-23 17:19:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071010000051KK00888