推 o5515334:thanks a lot 10/29 19:24
※ 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."
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