probability!thanks!

Another method for consideration:
First, we have to consider which rank does the selected girl child belong to, i.e. 排行第幾.
If she is the eldest (with probability = 1/3) in the family, definitely she does not have an older sister.
If she is the second (with probability = 1/3) in the family, then there is one event that she can have an older sister:
The eldest child is a girl → probability = 1/2
If she is the youngest (with probability = 1/3) in the family, then there are three events that she can have an older sister:
1) The eldest child is a girl but the second is a boy → probability = 1/4
2) The eldest child is a boy but the second is a girl → probability = 1/4
3) Both the eldest and second children are girls → probability = 1/4
So total probability is 3/4.
Integrating all possibilities above, we have the overall probability as:
(1/3) × (1/2) + (1/3) × (3/4) = 5/12.
註: 這題基本上同選出的 child 係男 or 女無關的. 如果 the question 被改成 "a child is selected at random and found to be a boy", 答案仍然是一樣.

唔識計. 所以最好就係用list.
3 個細路有8 種可能性.

1. GGG
2. GGB
3. GBG
4. GBB
5. BGG
6. BGB
7. BBG
8. BBB

提目given 係個女, 即係Case 1 -7, 一定唔係case 8.
所以係7個case 既其中一個.
如果係case 1, 抽出既係大女,係1/3. 係case 2,3 係 1/2, case 4, 係1/1.
所以..... 答案係
1/3 * 1/7 + 1/2 * 1/7 + 1/2 * 1/7 + 1/1 * 1/7

2007-02-02 13:56:03 補充：
上兩個都錯. 因為提目已經話, 已經抽出一個女. 所以唔應該考慮抽出女既prob. 抽出女既prob係冇關既.呢個係prior prob, 同 post prob 既分別. 你睇我既解,你就知, 你地將 3個仔既case 既prob 都計埋落去. 咁咪錯囉.

let B : boy, G:girl
All combinatioons
BBB BBG BGB BGG GBB GBG GGB GGG
Probability to pick a girl
=1/8*0+1/8*1/3+1/8*1/3+1/8*2/3+1/8*1/3+1/8*2/3+1/8*2/3+1/8*1
=1/8(12/3)
=1/2
the probability that pick a girl and she has an older sister
=1/8*0+1/8*0+1/8*0+1/8*2/3*1/2+1/8*0+1/8*2/3*1/2+1/8*2/3*1/2+1/8*1*2/3
=1/8(1/3+1/3+1/3+2/3)
=5/24
So the reqiured probability
=(5/24)/(1/2)
=10/24
=5/12

2007-02-02 13:43:17 補充：
BGB 表示最大的孩子B,第二大的孩子G,最小的孩子B

2007-02-02 15:34:56 補充：
最後用到公式:P(A|B)=P(A)/P(B)不過我估你本書都有

Assume probability of a boy and girl is 1/2,
so the probability that she has an older sister is:
1/2*1/2*1/2
=1/8