Four universities - 1, 2, 3, and 4 - are participating in a holiday basketball tournament. In 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 event that 1 wins the tournament. Then List outcomes in A.(C) Let B denote the event that 2 gets into the championship game. Then List outcomes in B.(D) What are the outcomes in A∪B and in A∩B? What are the outcomes in A'

Part A. {S} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Part B. {A} = {1324, 1342, 1423. 1432}

Part C. {B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

Part D.

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

(A∩B) = ∅

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

Step-by-step explanation:

A. All possible outcomes

There are four teams, each play a semi final where 1 and 2 plays against each other while 3 and 4 plays against each other. Winner of the first semi final can be either 1 or 2 therefore they both can not be in the championship game or in the losers game at the same time same goes for the other semi final.

Using this explanation (1324 denotes: 1 beats 2 and 3 beats 4 in first-round games and then 1 beats 3 and 2 beats 4), All possible outcomes are

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

B. Event A in which 1 wins the tournament

From {S} we only have to write the outcomes in which 1 is the first number in 4digit combinations given in part A

{A} = {1324, 1342, 1423. 1432}

C. Event B in which 2 gets into championship game

From {S} we only have to write the outcomes in which 2 is either the first or second digit in 4digit combinations given in part A

{B} = {2314, 2341, 2413, 2431, 3214, 3241, 4213, 4231}

D. Outcomes in (A∪B), (A∩B) and A'

I. (A∪B)

(A∪B) means A union B therefore all we have to do is combine all the members of A and B

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

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

II. (A∩B)

(A∩B) means A intersection B in which we have to find the common members of A and B. If there are no common members then the result of (A∩B) is a null set.

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

(A∩B) = ∅

III. {A'}

A' means A compliment, in other words it can be described as all the possible outcomes that are not part of A. So all we do is to subtract outcomes of A from the total possible outcomes S

{A'} =  {S} - {A}

{A'} = {1324, 1342, 1423, 1432, 2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231} - {1324, 1342, 1423. 1432}

{A'} = {2314, 2341, 2413, 2431, 3124, 3142, 3214, 3241, 4123, 4132, 4213, 4231}

a=24; b=12

step-by-step explanation:

a/b=2: 1

a=2b

36/(1+2)=12

b=12

a=24