Question

score on last try: 0 of 1 pts. see details for more. > next question get a similar question you can retry this question below a survey of athletes at a high school is conducted, and the following facts are discovered: 4% of the athletes are football players, 41% are basketball players, and 2% 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 = 40 x % (please enter your answer as a percent) submit question

Explanation:

Step1: Use the inclusion - exclusion principle

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. The formula for $P(F\cup B)$ is $P(F\cup B)=P(F)+P(B)-P(F\cap B)$.

Step2: Substitute given values

We know $P(F) = 41\%$, $P(B)=4\%$, and $P(F\cap B) = 2\%$. Then $P(F\cup B)=41 + 4- 2$.

Step3: Calculate the result

$P(F\cup B)=43\%$.

Answer:

43%