a box contains a mixture of washers of various sizes. there are 50 of type A, 30 of type B and 20 of type C. In the process of sorting them we take handfuls of 20 washers at a time. what is the probability that in the first handful we have:
a) exactly 20 type of A washers
b) at least 5 type C washers
c) No type A washers
數學問題一問
2007-09-29 7:22 pm
回答 (2)
2007-09-29 8:41 pm
✔ 最佳答案
上面那位朋友的part b 不對而且這類問題用Combination的方法去諗會比較好
a.
50C20 / 100C20
分母代表無任何限制下，在100個washers中抽出20個的組合方式
分子就是根據題目要求，要抽20個 type a 的washer的組合方式，因為total有50個type A, 所以是 50C20
b.
1- [ ( 80C20 + 80C19 * 20C1 + 80C18 * 20C2 + 80C17 * 20C3 + 80C16 * 20C4) / 100C20 ]
= 0.3647 (round to 4 d.p.)
At least 類的問題通常都是用 「1 減去something」的方法處理，題目要求at least 5 type C，就用 1 減 0,1,2,3和4個type C 的可能，餘下來就會是at least 5個typeC
其實邏輯上都要諗下個答案合唔合理，0.3647都好合理，因為100個中有20個係 type C，比例上都唔算太少，所以計出黎的答案唔應該細得太緊要
假如100個中只有 6 個type C，而又要抽中at least 5 type C，咁個答案就會好細好細
c. 50C20 / 100C20
因為type A一定不能被抽中，所以只有在餘下的type B和C (30+20) 中抽20個出黎，所以分子是 50C20
2007-10-02 10:09:45 補充：
我的補充過了字數上限，已e-mail給你
參考: 自己
2007-09-29 8:02 pm
a. exactly 20 type A washers
the probability of the first is 50/100, the second is 49/99, third=48/98..
The probability of getting all 20 type A washers is therefore
50*49*48....31/(100*99*....81)
=50!/30! /(100!/80!)
= 148/1683150111
b. at least 5 type C
20*/100 * 19/99 * 18/98 * 17/97 *16/96
=20!/15! / (100!/95!)
=323/1568490
c. No type A
50/100 * 49/99 * ... 31/81
=50!/30! / (100!/80!)
=148/1683150111
the probability of the first is 50/100, the second is 49/99, third=48/98..
The probability of getting all 20 type A washers is therefore
50*49*48....31/(100*99*....81)
=50!/30! /(100!/80!)
= 148/1683150111
b. at least 5 type C
20*/100 * 19/99 * 18/98 * 17/97 *16/96
=20!/15! / (100!/95!)
=323/1568490
c. No type A
50/100 * 49/99 * ... 31/81
=50!/30! / (100!/80!)
=148/1683150111
收錄日期: 2021-04-13 13:39:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070929000051KK01259