Question

a survey of athletes at a high school is conducted, and the following facts are discovered: 17% of the athletes are football players, 24% are basketball players, and 7% 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

The formula for $P(A\cup B)$ is $P(A)+P(B)-P(A\cap B)$. Let $A$ be the event that an athlete is a football - player and $B$ be the event that an athlete is a basketball - player. We know that $P(A) = 17\%=0.17$, $P(B)=24\% = 0.24$ and $P(A\cap B)=7\%=0.07$.

Step2: Substitute the values into the formula

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.17 + 0.24-0.07$.

Step3: Calculate the result

$0.17+0.24 - 0.07=0.34$.

Answer:

34%