If a soccer team won 60% of the games in which its players scored 5 or more goals, what is the probability of choosing a game in in which the team scored 0-4 goals and won the game? The season so far is they have won 3 games scoring 0-4 goals, won 3 games scoring 5 or more goals and lost 4 games scoring 0-4 goals. Answer supposed to be 25% but cannot get that answer
To solve this problem, we need to break it down into two steps:
Step 1: Calculate the probability of choosing a game in which the team scored 5 or more goals and won the game.
Step 2: Calculate the probability of choosing a game in which the team scored 0-4 goals and won the game.
Step 1: Calculate the probability of choosing a game in which the team scored 5 or more goals and won the game.
From the given information, we know that the team won 60% of the games in which its players scored 5 or more goals. Let's call this probability P(A), where A represents the event "the team scored 5 or more goals and won the game."
P(A) = 0.60
Step 2: Calculate the probability of choosing a game in which the team scored 0-4 goals and won the game.
To find the probability of this event, P(B), we need to subtract the probability of choosing a game in which the team scored 5 or more goals and won the game from 1 (since these two events are mutually exclusive).
P(B) = 1 - P(A)
P(B) = 1 - 0.60
P(B) = 0.40
The requested probability is the probability of choosing a game in which the team scored 0-4 goals and won the game. Since this probability is P(B) = 0.40, it is not equal to 25%.
Therefore, the answer provided (25%) does not correspond with the given information.