Solution to Exercise 2.1.

  Problem Statement: 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 $\cal S$.

Solution: We could list all 4! = 24 permutations of $\{ 1, 2, 3, 4 \}$, but not all permutations are possible representations of outcomes in the given coding system. It is not possible for 1 and 2 to be in the first or second pair of digits, nor for 3 and 4, since these would have played in the first round. Breaking things down somewhat systematically, here's the outcomes where 1 wins its first round game:

$$1324, \; 1342, \; 3124, \; 3142 , \; 1423, \; 1432, \; 4123, \; 4132 .$$

Here's the outcomes where 2 wins its first game:

$$2314, \; 2341, \; 3214, \; 3241 , \; 2413, \; 2431, \; 4213, \; 4231 .$$

That makes 16 outcomes in the sample space.

(b) Let A denote the event that 1 wins the tournament. List outcomes in A.

Solution:

$$A \; = \; \{ 1324, \; 1342, \; 1423, \; 1432 \} .$$

(c) Let B denote the event that 2 gets into the championship game. List outcomes in B.

Solution:

$$B \; = \; \{ 2314, \; 2341, \; 3214, \; 3241 , \; 2413, \; 2431, \; 4213, \; 4231 \} .$$

(d) What are the outcomes in $A \cup B$ and in $A \cap B$? What are the outcomes in $A^{\prime}$?

Solution:

$$A \cup B \; = \; \{ 1324, \; 1342, \; 1423, \; 1432, \; 2314... ...341, \; 3214, \; 3241 , \; 2413, \; 2431, \; 4213, \; 4231 \} .$$

$$A \cap B \; = \; \emptyset .$$

$$A^{\prime} \; = \; \{ 2413, \; 2431, \; 4213, \; 4231 \; 23... ...341, \; 3214, \; 3241 , \; 2413, \; 2431, \; 4213, \; 4231 \} .$$