问题及解答

Four universities -- 1,2,3 and 4 -- are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4. Then the two winners will play for the championship, and the two losers will also play. One possible outcome can be denoted by 1324 (1 beats 2 and 3 beats 4 in first round games, and then 1 beats 3 and 2 beats 4).

A. List all outcomes in $\mathcal{S}$.
B. Let $A$ denote the event that 1 wins the tournament. List outcomes in $A$.
C. Let $B$ denote the event that 2 gets into the championship game. List outcomes in $B$.
D. What are the outcomes in $A\cup B$ and in $A\cap B$? What are the outcomes in $A^{c}$?
 

Reference:

Jay L. Devore, Probability and Statistics, For Engineering and The Sciences (Fifth Edtion)

我们用 a,b,c,d 代表四个学校的球队. 那么这四个字母的任何一个排列就代表了一个比赛结果.

比如: cadb 表示第一轮比赛: c 战胜 d, a 战胜 b; 第二轮比赛 c 战胜 a 获得冠军, a获得亚军; d 战胜 b 获得季军.

用图表示如下:

c--\
      |---> c---\
d--/               |
                        ====> c

a--\               |
      |---> a---/
b--/

另一方面, a,b,c,d 的任意两个不同排列代表的比赛结果是不同的. 

因此, 总的比赛结果有 $P_4=4!=24$ 种.

但是这里题目中限定了第一轮是 1 与 2 比赛, 3 与 4 比赛. 因此, 不会出现下列八种情形.

1234, 1243
2134, 2143
3412, 3421
4312, 4321

因此, 样本空间为 $\{1,2,3,4\}$ 的全排列空间去掉上面八个元素.  也即

\[\mathcal{S}=\{ 1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231\}\]