拜託高手解答數學的機率計算問題！
2014-03-29 5:42 am
A fleet of nine taxis is to be dispatched to three airports in such a way that three go to airport A, five go to airport B, and one goes to airport C. (a) In how many distinct ways can this be accomplished? (b) If exactly one of the taxis is in need of repair, what is the probability that it is dispatched to airport C? (c) If exactly three of the taxis are in need of repair, what is the probability that every airport receives one of the taxis requiring repairs?
回答 (5)
2014-03-29 8:09 am
✔ 最佳答案
(a)Out of 9 taxis, 1 goes to airport C. (9C1)
Out of the rest 8 taxis, 5 go to airport B. (8C5)
The rest 3 taxis go to airport A. (3C3)
No. of ways this can be accomplished
= 9C1 x 8C5 x 3C3
= 9 x (8!/5!3!) x 1
= 504
(b)
When the event happens :
The taxi needed repair is dispatched to airport C. (1C1)
Out of the rest 8 taxis, 5 go to airport B. (8C5)
The rest 3 taxis go to airport A. (3C3)
The required probability
= (1C1 x 8C5 x 3C3)/ (9C1 x 8C5 x 3C3)
= 1C1 / 9C1
= 1/9
(c)
When the event happens :
The 3 taxis needed repair go the 3 different airports. (3P3)
Out of the rest 6 taxis, 4 go to airport B. (6C4)
The rest 2 taxis go to airport A. (2C2)
The required probability
= [3P3 x 6C4 x 2C2]/ 504
= [3! x (6!/4!2!) x 1] / 504
= [6 x 15 x 1] / 504
= 5/28
參考: 土扁
2014-05-30 5:16 pm
2014-04-27 11:56 pm
2014-04-25 3:47 pm
2014-04-24 2:43 pm
收錄日期: 2021-04-13 20:20:55
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https://hk.answers.yahoo.com/question/index?qid=20140328000010KK05079