Question

a survey of athletes at a high school is conducted, and the following facts are discovered: 39% of the athletes are football players, 25% are basketball players, and 22% 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? probability = % (please enter your answer as a percent)

Explanation:

Step1: Recall the formula for the probability of the union of two events

Let $P(F)$ be the probability of being a football - player, $P(B)$ be the probability of being a basketball - player, and $P(F\cap B)$ be the probability of being both a football and a basketball player. The formula for $P(F\cup B)$ is $P(F\cup B)=P(F)+P(B)-P(F\cap B)$.

Step2: Substitute the given values

We are given that $P(F) = 0.39$, $P(B)=0.25$, and $P(F\cap B)=0.22$.
$P(F\cup B)=0.39 + 0.25-0.22$.

Step3: Calculate the result

$P(F\cup B)=0.39+0.25 - 0.22=0.42$.

Answer:

42%