Re: [問題] 請教一個機率問題

看板CFAiafeFSA (精算師/基金經理人/銀行家)作者monzo時間17年前 (2008/09/05 21:04)推噓1(1推 0噓 2→)

留言3則, 2人參與討論串2/2 (看更多)

※ 引述《linki (111111)》之銘言： : 這是actex study manual module 3 poisson process的example 3.20 : 根據3.19的題意延伸 : p(wating time of the claim from Texas : < wating time of the claim from Missouri)=0.6 : 也就是claim from Texas會比Missouri早發生的機率為0.6 : 而3.20問到 : For the insurance company in the preceding example: : (a) Find the probablity that three claims from Texas are received before the : first claim from Missouri. : (b) Find the probablity that three claims from Texas are received before the : third claim from Missouri. : (a)是比較沒問題，就是發生的前三件事都是from Texas : 所以答案是0.6x0.6x0.6=0.216 : (b)我解讀題意是在第三次的案件來自Missouri前，發生三次案件來自Texas : 不知道有沒有解讀錯(不好意思，小弟英文不是很好) : 我的想法是用負二項分配來解題，也就是把claim from Missouri當成功(p=0.4) : 也就是利用負二項分配計算在第三次成功前，有三次失敗的機率 : c5取3x0.4^3x0.6^3=0.13824 : 但解答出來卻不是如此 : 想請教各位大大是我對題目解讀錯誤 : 還是觀念錯了 : 謝謝各位 : 附上(b)之解答 : Consider the first five claims received. Three claims from Texas will be : received before the third claim from Missouri if and only if three or more : of the first five claims come from Texas. The number of claims from Texas in : the first five claims received follows a binomial distribution with n=5 and : p=0.6 : The probability that three or more of the first 5claims came from Texas is : p(3)+p(4)+p(5)=0.6826 假設O是from Texas X是from Missouri (OOOXX)X p(3)=C5取3*(.6)^3*(.4)^2=0.3456 (OOOOX)XX p(4)=C5取4*(.6)^4*(.4) =0.2592 (OOOOO)XXX p(5)=C5取5*(.6)^5 =0.0777 --------------------------------------------- 0.6825 他解答應該是這個意思吧不過我看不懂題目為何這樣@@ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 219.84.16.110