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Chapter 2: Q1E (page 56)

Four universities鈥�1, 2, 3, and 4鈥攁re 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鈥�?

Short Answer

Expert verified

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

01

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.

02

Providing the elements of a sample space

a.

The sample space for the provided scenario is as follows,

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

03

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,

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

04

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,

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

05

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,

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

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