※ 引述《o5515334 (無)》之銘言： : 如果是跟統計軟體有關請重發文章 : 如果跟論文有關也煩請您重發文章 : 文章類別是為了幫助大家搜尋資料與解答，造成不便之處請見諒 : 題目如下 : (1). 一副牌共52張，4個花色，每個花色有13張牌，現在抽取2張且抽後不放回 : 試問第一張為10且第二張是5或6的機率為何? : 我的作法是這樣 [( C4取1 )*( C8取2 )]/(C52取2)=0.024 : 這樣做法正確嗎? : (2). 另外在複迴歸分析中，若X1變數的B值顯著，指在母體中X1變數的B值大體上不會是0 : 這句話有錯嗎? : (3). 還有樂透包牌的問題(42選6)，若包8個號碼，中頭獎的機率應該怎麼算呢? : (4). 最後是關於poisson分配，為什麼在極短時間間隔或區域內， : 特定事件發生超過一次之機率要忽略不計? 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." : 謝謝熱心回答的人~!感激不盡! -- Kraft Durch Freude~ --

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