Question

a survey of athletes at a high school is conducted, and the following facts are discovered: 60% of the athletes are football players, 13% are basketball players, and 6% 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

We use the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, where $A$ is the event of being a football - player and $B$ is the event of being a basketball - player.

Step2: Identify the given probabilities

Let $P(A) = 60\%=0.6$, $P(B)=13\% = 0.13$, and $P(A\cap B)=6\%=0.06$.

Step3: Substitute the values into the formula

$P(A\cup B)=0.6 + 0.13-0.06$.
$P(A\cup B)=0.67$.

Answer:

67%