There are 3 doors.Each door can be opened by one key only and no two doors can be opened by the same key.The keys of these 3 doors together with 7 other keys are put in a drawer.If Jimmy chooses 3 keys ramdomly from the drawer,find the probability that Jimmy opens each doors at one trial.
ans:1/720
F.5 probability question
2011-12-15 10:02 am
回答 (2)
2011-12-15 10:27 am
✔ 最佳答案
(Toopen the first door, the correct key must be chosen from the 10 keys.Probability = 1/10)(To open the second door after the first door is opened, the correct key mustbe chosen from the remained 9 keys. Probability = 1/9)
(To open the third door after the first and second doors are opened, thecorrect key must be chosen from the remained 8 keys. Probability = 1/8)
The required probability
= (1/10) x (1/9) x (1/8)
= 1/720
參考: micatkie
2011-12-20 8:54 pm
i have different opinion according to the question.
as it doesn't specify the "order" of doors to be opened by the key,
so the probability of first key can open the door is 3 out of 10 = 3/10
so the probability of second key can open the door is 2 out of 9 = 2/9
so the probability of last key can opend the door is 1 out of 8 = 1/8
the probability at one trial = 3/10 x 2/9 x 1/8 = 1/120
however if the "order" of doors to be opened in sequence, the probability at one trial is believed 1/720.
by the way, according to the question, i would choose 1/120 as my answer
as it doesn't specify the "order" of doors to be opened by the key,
so the probability of first key can open the door is 3 out of 10 = 3/10
so the probability of second key can open the door is 2 out of 9 = 2/9
so the probability of last key can opend the door is 1 out of 8 = 1/8
the probability at one trial = 3/10 x 2/9 x 1/8 = 1/120
however if the "order" of doors to be opened in sequence, the probability at one trial is believed 1/720.
by the way, according to the question, i would choose 1/120 as my answer
收錄日期: 2021-04-13 18:25:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111215000051KK00063