一个很紧急的问题。。。

Dicardo

a box contain 5 red ball, 4 blue ball, 3 yellow ball.
three balls are selected at random.

Find that the probability that all the threee ball are different colours.

为什么不是 1/5 x 1/4 x 1/3 呢？因为那到red , blue, yellow 的机会分别是 1/5 1/4 1/3 嘛。。
为什么是 (5C1 x 4C1 x 3C1) / (12C3) ?

harry_lim

5 red ball, 4 blue ball, 3 yellow ball

你要找出3个ball里面要different colour...
1)红，蓝，黄
2)蓝，红，黄
3)黄，红，蓝

n(S) = 12
n(Red) = 5
n(Blue) = 4
n(Yellow) = 3

P(R) = 5/12
P(B) = 4/12
P(Y) = 3/12

P(different colour) = (5/12 × 4/11 × 3/10) + (4/12 × 5/11 × 3/10) + (3/12 × 5/11 × 4/10)
                 = 3/22
(我的答案对不对？   )
我很久没动probability了

Dicardo

原帖由 harry_lim 于 28-8-2009 11:55 AM 发表
5 red ball, 4 blue ball, 3 yellow ball

你要找出3个ball里面要different colour...
1)红，蓝，黄
2)蓝，红，黄
3)黄，红，蓝

n(S) = 12
n(Red) = 5
n(Blue) = 4
n(Yellow) = 3

P(R) = 5/12
P(B) = ...

对哦..
我最讨厌这一课了...:@

harry_lim

我的答案对不对的？

Dicardo

原帖由 harry_lim 于 28-8-2009 12:05 PM 发表
我的答案对不对的？

对哦。。。

乙劍真人

好像是 3/11 哩..

harry_lim

原帖由 Dicardo 于 28-8-2009 12:17 PM 发表


对哦。。。

其实我是用很简单方法做的
因为我的。。。头脑简单

harry_lim

我算的。。。。。。。
P(different colour) = (5/12 × 4/11 × 3/10) + (4/12 × 5/11 × 3/10) + (3/12 × 5/11 × 4/10)
                 = 3/22


3/22是对的吗？？？
到底是3/11还是3/22?

Dicardo

嘻嘻。。 看错了。。 是3/11

harry_lim

原帖由 Dicardo 于 28-8-2009 12:24 PM 发表
嘻嘻。。 看错了。。 是3/11

给你炸到下

Dicardo

为什么要用5C1 ? 我还是不明白

乙劍真人

Harry 漏了另外三个可能性！

你要找出3个ball里面要different colour...
1)红，蓝，黄
2)蓝，红，黄
3)黄，红，蓝
4)红，黄，蓝
5)蓝，黄，红
6)黄，蓝，红

n(S) = 12
n(Red) = 5
n(Blue) = 4
n(Yellow) = 3

P(R) = 5/12
P(B) = 4/12
P(Y) = 3/12

P(different colour) = (5/12 × 4/11 × 3/10) + (4/12 × 5/11 × 3/10) + (3/12 × 5/11 × 4/10) +
                                   (5/12 x 3/11 x 4/10) + (4/12 x 3/11 x 5/10) + (3/12 x 4/11 x 5/10)

                                = 6/22
                                = 3/11

最快的做法：

(5C1 X 4C1 X 3C1) / (12C3) = 3/11

乙劍真人

原帖由 Dicardo 于 28-8-2009 12:30 PM 发表
为什么要用5C1 ? 我还是不明白

5 4 3

5C1 是说五粒里任选一粒

Dicardo

原帖由 乙劍真人 于 28-8-2009 12:33 PM 发表


5 4 3

5C1 是说五粒里任选一粒

5 选1 ..
也可以 1/5 啊。。。

harry_lim

原帖由 乙劍真人 于 28-8-2009 12:31 PM 发表
Harry 漏了另外三个可能性！

你要找出3个ball里面要different colour...
1)红，蓝，黄
2)蓝，红，黄
3)黄，红，蓝
4)红，黄，蓝
5)蓝，黄，红
6)黄，蓝，红

n(S) = 12
n(Red) = 5
n(Blue) = 4
n(Yell ...

原来我还少算

harry_lim

我做probability的题目。。。比较不会用nPr,nCr的方法算的

乙劍真人

原帖由 Dicardo 于 28-8-2009 12:34 PM 发表

5 选1 ..
也可以 1/5 啊。。。

嗯

虽然说每粒得到的概率是 1/5，但是里面有五粒哦
那五粒用 permutation 来排你就会发现不能用 1/5 了..

乙劍真人

原帖由 Dicardo 于 28-8-2009 12:34 PM 发表

5 选1 ..
也可以 1/5 啊。。。

当题目问五粒选一粒，其概率是 1/5 没错，但是这个题目是问三个不同颜色里选三粒

5 4 3

你五粒里选一粒是 1/5，别忘了另外四粒也有可能被选中哦

darksider

P ( 3 balls with different colours = a) = events a / total events
Therefore P(a) =  [5C1 x 4C1 x 3C1] / [ 12C3 ]
                         = [5!/(4!1!) x 4!/(3!1!) x 3!/(2!1!)] / [12!/(9!3!)]
                         = 3/11

Let's recall permutation. Permutation helps us to find out how many ways can a given objects be allocated to a certain space.

Let say now I have 2 chairs, M and N. 3 people, A , B and C, want to sit on them.

Initially, chair M can possibly be sitted by either A , B or C.
When chair M is sitted, chair N can only be sitted by either one of the remaining 2 people.
So the possible different arrangements = 3 x 2 = 6 as shown below
AB AC BC
BA CA CB  
The general formula of permutation is nPk = n! / (n-k)! , n = number of people and k = number of chairs

But now I want to find out, how ways can the 3 people be chosen into 2 chairs (take note that I don't care about the arrangements, I just care about them being chosen to sit on those chairs)

So now, we know that there are 6 different arrangments.
Since we care about them being chosen, we divide the 3P2 by 6 (because I don't care about the different arrangements)
Hence we get 3P2/6 = 3P2/3! = [ 3!/(3-2)!] / 3! = 3
So there are only 3 ways which they can be chosen, ie , AB, BC and CA.
The general formula of combination is nCk = n!/(n-k)!(k)! = nPk/k!

So back to your question.
You have 5 red balls and you only want to choose one of them.
So the number of ways which you can pick 1 red ball out of 5 = 5C1
You have 4 blue balls and you again want to pick only one.
So the number of ways = 4C1
You have 3 yellow balls and you pick only one
So the number of ways which you can pick them = 3C1

P(3 balls with different colours are picked = a) = n(a) / total events
                                                                         = (5C1 x 4C1 x 3C1)/ (12C3)
                                                                         = 3/11

Total events is such that you pick 3 balls out of 12 balls which is equal to 12C3.

Hopefully you understand.

[ 本帖最后由 darksider 于 28-8-2009 05:39 PM 编辑 ]

Log

3P2/6=I not equal to 3???