# Quod Erat Demonstrandum

## 2008/07/04

### Just answer an applied mathematics question

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 5:26 下午

Just answer an applied mathematics question from a F.6 student.

Book 4A Ex 2(a) No.19.

An urn consists of n tickets labelled 1, 2, 3, …, n. Tickets are drawn one by one without replacement. What is the probability that in these n draws, there is at least one for which the number on the ticket corresponds to the number of draw.

The following is a solution given by the student.

1. Denominator is n! because there are n tickets lining up for n size of queue.

2. For P(1), there are nC1 ways for the 1 ticket which is corresponding to no. of draw and the rest n-1 tickets are lining up for n-1 size of queue. So P(1)=nC1(n-1)!/n!.

3. Similarly, P(2)=nC2(n-2)!/n!, P(3)=…, P(n)=nCn/n!

5. Hence the answer is [nC1(n-1)!+nC2(n-2)!+nC3(n-3)!+…+nCn]/n! = 1+1/2!+1/3!+…+1/n!

Obviously, the answer is not correct because it is greater than 1. So, what’s wrong?

http://johnng.inscyber.net/apm/pa_expand_union.doc

## 1 則迴響 »

1. oic~thx
看埋個入信封問題的document終於明白佢個(-1)^(n+1)係點走出來…

迴響 由 Sit — 2008/07/04 @ 11:05 下午 | 回應