.一袋中有3個紅球2個白球與3個黑球，隨機抽取4個球，令X表示紅球的個數，Y表示白球的個數，求:
(1)X與Y的聯合機率分配
答案是
(x,y) f(x,y)
(0,1) 2/70
(0,2) 3/70
(1,0) 3/70
(1,1) 18/70
(1,2) 9/70
(2,0) 9/70
(2,1) 18/70
(2,2) 3/70
(3,0) 3/70
(3,1) 2/70
有關於聯合機率分配表的答案我不太董怎麼解
比如說 3,0 為什麼會是3/70
拜託大大講解清晰點~~
有關於聯合機率分配 不太懂
2013-05-01 6:30 pm
回答 (3)
2013-05-01 7:45 pm
✔ 最佳答案
一袋中有3個紅球2個白球與3個黑球，隨機抽取4個球，令X表示紅球的個數，Y表示白球的個數，求:(1)X與Y的聯合機率分配
==========================
全部有八個球
取四個
所以是 c 8 取 4 = 70
2013-05-01 11:49:04 補充：
(3,0) 表示紅球3個，白球0個 => 黑球一個
所以
紅球 c 3 取 3 = 1
白球 c 2 取 0 = 1
黑球 c 3 取 1 = 3
1*1*3 = 3
P(3,0) = 3/70
2013-05-01 14:24:47 補充：
補充消失了?
2013-05-01 14:25:55 補充：
又出現了....
2013-05-01 9:15 pm
完全聽不懂 都說了要詳細點
2013-05-02 7:15 am
首先抽出的組合數是C(8,4) = 70
f(0,1) = C(3,0)C(2,1)C(3,3)/70 = 2/70
f(0,2) = C(3,0)C(2,2)C(3,2)/70 = 3/70
f(1,0) = C(3,1)C(2,0)C(3,3)/70 = 3/70
f(1,1) = C(3,1)C(2,1)C(3,2)/70 = 18/70
f(1,2) = C(3,1)C(2,2)C(3,1)/70 = 9/70
f(2,0) = C(3,2)C(2,0)C(3,2)/70 = 9/70
f(2,1) = C(3,2)C(2,1)C(3,1)/70 = 18/70
f(2,2) = C(3,2)C(2,2)C(3,0)/70 = 3/70
f(3,0) = C(3,3)C(2,0)C(3,1)/70 = 3/70
f(3,1) = C(3,3)C(2,1)C(3,0)/70 = 2/70
f(0,1) = C(3,0)C(2,1)C(3,3)/70 = 2/70
f(0,2) = C(3,0)C(2,2)C(3,2)/70 = 3/70
f(1,0) = C(3,1)C(2,0)C(3,3)/70 = 3/70
f(1,1) = C(3,1)C(2,1)C(3,2)/70 = 18/70
f(1,2) = C(3,1)C(2,2)C(3,1)/70 = 9/70
f(2,0) = C(3,2)C(2,0)C(3,2)/70 = 9/70
f(2,1) = C(3,2)C(2,1)C(3,1)/70 = 18/70
f(2,2) = C(3,2)C(2,2)C(3,0)/70 = 3/70
f(3,0) = C(3,3)C(2,0)C(3,1)/70 = 3/70
f(3,1) = C(3,3)C(2,1)C(3,0)/70 = 2/70
收錄日期: 2021-04-27 19:21:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130501000015KK01078