※ 引述《[email protected] (OK Computer.......)》之铭言： > Suppose two teams are playing a series of games, each of which is > independently won by team A with probability p and by team B with > probability 1-p. The winner of the series is the first team to win > i games. > 1).If i= 4,find the probability that a total of 7 games are played. > Also show that this probability is maximized when p=1/2. P(7 games are played) = P(前6局3-3比) = C(6,3)p^3(1-p)^3 显然最大值发生於 p=1/2. > 2).Find the expected number of games that are played when > (a).i=2 , (b).i=3 i=2, 最少比两场: pp+qq, where q=1-p 最多比3场: 2pq i=3, 最少比3场: ppp+qqq 最多比5场: 6(pq)^2 比4场的机率很显然. > 谢谢帮忙!!! 若一般的 i, 才是比较复杂的问题. 就上列问题而言, 只 是直接计算而已. -- 嗨!  你好! 祝事事如意, 天天 happy! 统计专业版, 需要你的支持! :) 无名小站 telnet://wretch.twbbs.org Statistics (统计方法讨论区) 盈月与繁星 telnet://ms.twbbs.org Statistics (统计：让数字说话) 成大计中站 telnet://bbs.ncku.edu.tw Statistics (统计方法及学理讨论区) 交大资讯次世代 telnet://bs2.twbbs.org Statistics (统计与机率) ★本文未经本人同意请勿转载; 回覆请勿全文引用, 请仅留下直接涉及部分。 -- 夫兵者不祥之器物或恶之故有道者不处君子居则贵左用兵则贵右兵者不祥之器非君子 之器不得已而用之恬淡为上胜而不美而美之者是乐杀人夫乐杀人者则不可得志於天下 矣吉事尚左凶事尚右偏将军居左上将军居右言以丧礼处之杀人之众以哀悲泣之战胜以 丧礼处之道常无名朴虽小天下莫能臣侯王若能守之万物将自宾天地相合以降甘露民莫 之令而自均始制有名名亦既有夫亦将知止知止可以不殆譬道之在天 163.15.188.87海