※ 引述《o5515334 (无)》之铭言： : 如果是跟统计软体有关请重发文章 : 如果跟论文有关也烦请您重发文章 : 文章类别是为了帮助大家搜寻资料与解答，造成不便之处请见谅 : 题目如下 : (1). 一副牌共52张，4个花色，每个花色有13张牌，现在抽取2张且抽後不放回 : 试问第一张为10且第二张是5或6的机率为何? : 我的作法是这样 [( C4取1 )*( C8取2 )]/(C52取2)=0.024 : 这样做法正确吗? : (2). 另外在复回归分析中，若X1变数的B值显着，指在母体中X1变数的B值大体上不会是0 : 这句话有错吗? : (3). 还有乐透包牌的问题(42选6)，若包8个号码，中头奖的机率应该怎麽算呢? : (4). 最後是关於poisson分配，为什麽在极短时间间隔或区域内， : 特定事件发生超过一次之机率要忽略不计? I assume you were talking about Poisson Process, since you mentioned "time" ==> P{N(h)>1} = o(h) By definition and letting u be the parameter, P{N(h)=0} = e^(-uh) = 1 - uh + o(h) P{N(h)=1} = uh*e^(-uh) = uh + o(h) (using Taylor's expansion, first order) Then, P{N(h)>1} = 1 - P{N(h)=0} - P{N(h)=1} = o(h) where o(h)/h -> 0 as h -> 0 (the decreasing rate of o(h) is faster than that of h) Accordingly, you can use the above to show that the rate of occurance of a Poisson Process in [t, t+h] is u, by letting h -> 0. This implies a (standard) Poisson Process has "unit jump" For random jump-size, you can refer to "Compound Poisson Process." : 谢谢热心回答的人~!感激不尽! -- Kraft Durch Freude~ --

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