Question

g A survey of athletes at a high school is conducted, and the following facts are discovered: 41% of the athletes are football players, 52% are basketball players, and 35% of the athletes play both football and basketball. An athlete is chosen at random from the high school: what is the probability that the athlete is either a football player or a basketball player

Answer 1

0.58 = 58% probability that the athlete is either a football player or a basketball player

Explanation:

We solve this question treating the probabilities as Venn's sets.

I am going to say that:

Event A: Athlete is a football player.

Event B: Athlete is a basketball player.

41% of the athletes are football players:

This means that
`P(A) = 0.41`

52% are basketball players

This means that
`P(B) = 0.52`

35% of the athletes play both football and basketball.

This means that
`P(A \cap B) = 0.35`

What is the probability that the athlete is either a football player or a basketball player

This is
`P(A \cup B)`, which is given by the following equation:

`P(A \cup B) = P(A) + P(B) - P(A \cap B)`. So

`P(A \cup B) = 0.41 + 0.52 - 0.35 = 0.58`

0.58 = 58% probability that the athlete is either a football player or a basketball player