Challenging probability

2010-05-03 11:06 am

n people go to a restaurant for dinner. Each puts an umbrella into a bin. When they leave, each of them randomly takes one umbrella.

Find the probability that:
None of the people get their own umbrellas.
Only 1 person get his own umbrella.
Only 2 people get their own umbrellas.
Only 3 people get their own umbrellas.
：
：
：
Only n–2 people get their own umbrellas.
All of the people get their own umbrellas.

Note: There is no need to find the probability that only n–1 people get their own umbrellas since its value should be 0. It is because if n–1 people get their own umbrellas, the reminding person must get his own umbrella.

更新1:

Hence find the expected number of the people who get their own umbrellas, i.e. find Σ(k = 0 to n) k P(n,k) directly from the result of P(n,k) , where P(n,k) is the probability that only k people get their own umbrellas.

更新2:

在http://hk.knowledge.yahoo.com/question/question?qid=7010050200056，nelsonywm2000竟然計算出無論任何n值，expected number of the people who get their own umbrellas總是恆為1。我很懷疑nelsonywm2000用錯了概念。

回答 (3)

天助

2010-05-10 2:48 am

✔ 最佳答案

exact k mathches (0<=k<=n)

number of sample space=n!
number of exact k mathches=f(k)
=C(n,k)*[ (n-k)!- C(n-k,1)*(n-k-1)!+ C(n-k,2)*(n-k-2)!-....]
=n!/[k!(n-k)!]*{ (n-k)!- (n-k)!/1!+ (n-k)!/2!-....]
=n!/k!*{ 1- 1/1!+ 1/2! - 1/3!+....+(-1)^(n-k)/(n-k)! }

so, p( exact k mathches)=f(k)/n!
=[ 1- 1/1!+ 1/2!- 1/3!+...+(-1)^(n-k)/ (n-k)!] /k!
for k=0,1,2,3,....,n

👍 0 👎 0

doraemonpaul

2010-05-17 3:22 pm

我的意思是藉著計算Σ(k = 0 to n) {k/k! [1 - 1/1! + 1/2! - 1/3! + ...... + (- 1)^(n - k)/(n - k)!]}去檢查expected number of the people who get their own umbrellas是否真正恆為1。

👍 0 👎 0

Jacob

2010-05-04 7:17 pm

P(exactly k matches)
= 1/k! [1/2! - 1/3! + 1/4! - ... + (-1)^(n-k) / (n-k)!]

When n→∞, P(exactly k matches) = e^(-1) / k!

2010-05-11 10:26:24 補充：
Expected value的確是1。

Let X be the number of ppl matches.
X = X_1 + X_2 + ... + X_n
where X_i = 1 if i-th person matches, X_i = 0 otherwise.

Now
P{X_i = 1} = P{i-th person matches} = 1/n
E[X_i] = 1*P{X_i = 1} + 0*P{X_i = 0} = 1/n

Hence
E[X] = E[X_1] + ... + E[X_n] = n*1/n = 1

👍 0 👎 0

收錄日期: 2021-04-30 14:21:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100503000010KK01008

檢視 Wayback Machine 備份