Question

Suppose you were a very wealthy individual who was interested in buying a soccer team. You were considering purchasing a team that tended to win 42% of their games, to lose 52% of their games, and to tie in 6% of their games.

1. In any one game, what is the probability that the team will either win or tie.

2. Suppose you go to see this team play in a five-game series, assume that success in each game is independent of success in any other game and the team plays all five games. What is the probability that the team loses their first game and then wins games 2,3,4, and 5?

- I need help determining which probability method to use, whether to add or multiply the numbers.

Answer 1

p = probability of team winning = 42% = 0.42      
q = probability of team losing = 52% = 0.52      
r = probability of team has a tie = 6% = 0.06      
A team can either win, lose or have a tie.      
Hence the three events are mutually exclusive      
       
1) To find P(team wins or has a tie)      
= P(team wins) + P(has a tie)      
= 0.42 + 0.06      
= 0.48      
P(team wins or has a tie) = 0.48      
       
2) P(team loses first game AND wins all the remaining 4 games)      
= P(team loses first game) AND P(team loses 2nd) AND P(team loses 3rd) AND P(team loses 4th) AND P(team loses 5th)      
= 0.52 x 0.42 x 0.42 x 0.42 x 0.42      
= 0.0162      
P(team loses first game AND wins all the remaining 4 games) = 0.0162