Question

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 Bdenote 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’?

Short answer

a. The outcomes are:

\(S = \left\{ {1324,3124,1342,3142,1423,1432,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

b. The possible outcomes of A are:

\(A = \left\{ {1324,1342,1423,1432} \right\}\)

c. The possible outcomes of B are:

\(B = \left\{ {2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

d. The possible outcomes are,

\(A \cup B = \left\{ {1324,1342,1423,1432,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

\(A \cap B = \phi \)

\(A' = \left\{ {3124,3142,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

Step-by-step solution

Given information

The number of universities is 4; they are 1, 2, 3, and 4.

In the first-round, 1 will play 2 and 3 will play 4. Then two winners and two losers will play respectively.

The possible outcome 1324 implies that 1 beat 2 and 3 beat 4 in the first round and then 1 beat 3 and 2 beat 4.

Providing the elements of a sample space

a.

The sample space for the provided scenario is as follows,

\(S = \left\{ {1324,3124,1342,3142,1423,1432,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

Providing the possible outcomes of A

b.

A represents the event that 1 wins the tournament.

Referring to the sample space of part a,

\(S = \left\{ {1324,3124,1342,3142,1423,1432,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

The outcomes in A will be the ones where 1 is at the beginning.

The possible outcomes of A is represented as,

\(A = \left\{ {1324,1342,1423,1432} \right\}\)

Providing the possible outcomes of B

c.

B represents the event that 2 gets into the championship game.

Referring to the sample space of part a,

\(S = \left\{ {1324,3124,1342,3142,1423,1432,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

The outcomes in B will be the ones where 2 is at any of the first two places.

The possible outcomes of B is represented as,

\(B = \left\{ {2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

List of the possible outcomes of union and intersection of events.

Referring to part b,

\(A = \left\{ {1324,1342,1423,1432} \right\}\)

Referring to part c,

\(B = \left\{ {2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

A union of two events A and B consists of all the outcomes that are either in A or B or in both events.

The possible outcomes in\(A \cup B\)is given as,

\(A \cup B = \left\{ {1324,1342,1423,1432,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

The intersection of two events A and B consists of all the outcomes that are present in both events.

The possible outcomes in\(A \cap B\)is given as,

\(A \cap B = \phi \)

This implies that are no possible outcomes in\(A \cap B\).

The complementary of an event A consists of all the outcomes that are not contained in A.

The possible outcomes in\(A'\)are given by,

\(A' = \left\{ {3124,3142,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)