Sample Spaces--Outcomes and Events

Question

The question I am working on is, "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 $S$.

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 $A'$?"

For part a), I got the set $S=\{1234, ~1432, ~1324, ~1342, ~2134, ~2431, ~2314, ~2143, ~3124, ~3421, ~3214, ~3142, ~4123, ~4321, ~4213, ~4132\}$ However, in the answer, 1234, 2134, 2143, 3421, and 4321 are not included in the sample space. And the ones I inadvertently omitted are 1423, 2341, 3241, 2413, and 4231.

Why didn't the answer key include the events that I did? From my understanding, the leading number in the string of digits is the ultimate winner, of both the preceding match to the championship and the championship itself. So, for instance, 1234 and 2134 would distinct events, because although 1 beat 3, 1 still didn't win the championship match in the 2134 as it did in 1234. Could someone possibly help me?

Answer 1

You’re overlooking the assumption that the first-round matchups were $1$ vs. $2$ and $3$ vs. $4$. The outcome $2134$ is therefore impossible, because it says that $2$ and $1$ were the first-round winners: they played each other in the first round, so they cannot both have won.

Suppose that the outcome is $abcd$. Then $a$ and $b$ were the first-round winners, and $c$ and $d$ were the first-round losers. Exactly one of $1$ and $2$ will be a first round winner, and the other will be a first-round loser, so one of $1$ and $2$ will be $a$ or $b$, and the other will be $c$ or $d$. Similarly, one of $3$ and $4$ must be a first-round winner and hence either $a$ or $b$, and the other must be a first-round loser and therefore either $c$ or $d$.

Any one of the four teams can be the overall winner, so $a$ can be $1,2,3$, or $4$. If $a$ is $1$ or $2$, $b$ must be $3$ or $4$, and if $a$ is $3$ or $4$, $b$ must be $1$ or $2$: the two first-round winners have to come from different first-round matchups. So far, then, we have eight possibilities:

$$13xx,14xx,23xx,24xx,31xx,32xx,41xx,\text{ and }42xx\;.\tag{1}$$

Suppose that the first-round winners were $2$ and $3$, and $3$ beats $2$ in the second round; then the outcome has the form $32xx$. The other two teams are $1$ and $4$, who were the first-round losers; either of them might win their second-round game, so the final outcome could be either $3214$ or $3241$. Similar reasoning applies to each of the possibilities in $(1)$, so we get a total of $16$ possible outcomes:

$$\begin{align*} &1324,1423,2314,2413,3124,3214,4123,4213\\ &1342,1432,2341,2431,3142,3241,4132,4231 \end{align*}$$

The four outcomes furthest to the left are those in which $1$ wins the tournament; the eight in columns $3,4,6$, and $8$ are the outcomes in which $2$ reaches wins in the first round and therefore gets into the championship game.

Answer 2

In S the first no. can be anything. the 2nd must be such that it wasn't opposing the first no. in the first round. The 3rd and 4th are the remaining no.s (can be in any order).

Though it's not asked for,

$ \left|{S}\right|=2^4=16 $

In A the first no. is 1. In B 2 is one of the 1st two no.s . Due to the fact that both 1 and 2 can't win in a head to head match,

$A\cap B= \varnothing $