概率問題

我的同學出的一個問題,看看大家懂不懂計?

巴士A10分鐘一班,巴士B15分鐘一班,你兩部都可以上,問你等巴士的平均時間是多少(expectation).

唔係就係12.5咩...

no...因為首先,你不是剛到時走左巴士.而且,最多都只會等10分鐘,個MEAN點會>10分鐘??

本篇文章已被 阿暪 於 Apr 7 2005, 15:49 編輯過

Can you choose to wait for the bus at anytime sin? If so then avg should be 5 mins....

Do the FIRST Bus A & B arrive at the same time?



↓ hehe i was thinking about that too.....

本篇文章已被 Pearltea 於 Apr 7 2005, 16:42 編輯過

要知道有沒有A與B同時到來的情況出現
如果這也是要自己計算的概率之一可真複雜

本篇文章已被 鮑鮑 於 Apr 7 2005, 16:37 編輯過

4.166666.............

3分鐘？

考慮到inspection paradox的話就要稍長一些。

4.166666 is assuming that the first bus comes at the same time, but consider that the lag of the first bus is random(evenly distributed).

For the 3 minutes, you have assumed that 6 bus A and 4 busB within 1 hour would mix up together and evenly distributed into 6 minutes 一班.

Therefore, both of them CAN BE a answer , but not the best.

I don't make sure the following:
If the lag is random, I think the mean of all situation is between 4.1666666 and 3.75. Is 3.854166666? But it seem median, not a mean.

如果空檔時段長短不一兼隨機分布的話，期望值就肯定比3分鐘長，但具體數字怎麼計我就不知道了。

機會率的話，假設巴士沒有誤點，
then：
巴士最早會於0秒後出現，
巴士最遲會於900秒後出現(15分鐘)，
所以，平均數是15/2
=7.5分鐘
或=450秒

The maximum waiting time should be 10 Mins ah...Bus A is every 10 mins..

For Guest, As the Expected time does not varies with the lag time(t) linearly, it is not 3.854166666.(In fact, it should be a parabola).

It is a question of Probability, its better to know the concept of Probability Density Function and Integration in order to solve the problem.

I have two solution for the answer. One considering the lag time and take average, one considering the probability density for a particular waiting time.

For Superdog, what is inspection paradox.....may you tell me amore about it......

Sorry for typing Engish, it is because I type Chinese Quite slow, and i want to type quite many things in this reply.....

補充一下:你到達巴士站的時間是純粹random, evenly distributed. 而巴士的班次是exactly 10 分鐘 和 15 分鐘一班,這應該和inspection paradox 不同.

本篇文章已被 阿暪 於 Apr 9 2005, 06:08 編輯過

除了本身就知道答案者，相信再沒有甚麼人有意見了。還請阿瞞兄盡快專心打answer & solution。我自己只去到guest的程度，沒有甚麼要講。
另外，這主題與上次元直兄的生牌死牌問題相類，似乎可以搬過去宛城。

Probability of bus A arriving at different time should be evenly distributed between 0 and 10 minutes.
Probability of bus B arriving at different time should be evenly distributed between 0 and 15 minutes.

Probability of bus A arrive between time x and x + dx = 1/10 * dx
Probability of bus B arrive between time x and x + dx = 1/15 * dx

Probability of bus A arrive after time x = (10 - x) / 10
Probability of bus B arrive after time x = (15 - x) / 15


For the probibility of you get bus A or bus B between time x and x + dx
= (A arrive AND B not arrive yet ) OR (B arrive AND A not arrive yet)
= 1/10*(15-x)/15*dx+1/15*(10-x)/10*dx
= (25 - 2x)/150 * dx

P(x)dx = (25 - 2x)/150 * dx
P(x) = (25 - 2x)/150

E[P(x)] = Integrate from 0 to 10 (x(25-2x)/150)dx
= 35/9


大家認為對不對???

本篇文章已被 阿暪 於 Apr 9 2005, 07:48 編輯過

兩輛車的出發時間是否相同？

本篇文章已被 ~黃昭~ 於 Jul 28 2005, 10:34 編輯過

每小時(=60分鐘)有10班車經過(6A, 4B), 所以是平均6分鐘一班.

沒關係/不知道,總之兩條巴士線路是不多不少10分鐘一班和15分鐘一班,剛才的一班走了多久是隨機(0-10)/(0-15).

其實你說的問題之前的回覆也有說過.

不太正確,那樣兩巴士到達時間的distribution已經不同了.而且我是問平均等候時間.

例如說,如果A 和 B 巴士都是10分鐘一班,那A 和 B 是五分鐘間隔(夾硬加起來在取平均) 和永遠同時間到達(其中一種分布)的平均等候時間已經不同了,何況我們要考慮所有不同的distribution?而且10分鐘和15分鐘根本不能這樣分布.

如果是A,B都是10分鐘一班,那應該是:

P(x) = x/10*(10-x)/10 + (10-x)/10*x/10
= (10x-x^2)/50

E[P(x)] = In 0,10 [P(x)]
= x^2/10 - x^3/150 | 0,10
= 10/3 = 3.33...





本篇文章已被 阿暪 於 Oct 19 2005, 10:29 編輯過

我笨笨.........我有點不明白....
樓主可以解釋一下這句soln.嗎?