Two baseball teams are tied for first place at the end of the season. According to tie breaking rules, one of the teams will be designated as "Team X" and the other as "Team Y." Team X will then choose between two scenarios.
Scenario 1: The tied teams will play two games in Team X's ballpark. Team X must win BOTH games to be declared the champion. Otherwise, Team Y is the champion.
Scenario 2: The tied teams will play one game in Team Y's ballpark. The winner of this game will be declared the champion.
(Refer to tie breaking scenarios 1 and 2.)
If Team X has a 75% chance of winning each game played in its own ballpark and a 50% chance of winning a game played in Team Y's ballpark, which scenario should Team X choose?
A. Scenario 1
B. Scenario 2
C. Neither scenario is more likely than the other to result in a championship for Team X according to these figures.
I think its C
A senario 1
To determine which scenario Team X should choose, we need to calculate the probabilities of winning the championship under each scenario.
In Scenario 1, Team X must win both games played in their own ballpark to be declared the champion. We are given that Team X has a 75% chance of winning each game played in their own ballpark. To calculate the probability of winning both games, we multiply the individual probabilities together: 0.75 * 0.75 = 0.5625 (or 56.25%).
In Scenario 2, Team X plays one game in Team Y's ballpark. We are given that Team X has a 50% chance of winning a game played in Team Y's ballpark. Therefore, the probability of winning the championship in this scenario is simply 0.5 (or 50%).
Comparing the probabilities, we see that the probability of winning the championship in Scenario 1 (56.25%) is higher than the probability in Scenario 2 (50%). Therefore, Team X should choose Scenario 1 in order to maximize their chances of winning the championship.
So the correct answer is A. Scenario 1.