Question: A survey of athletes at a high school is conducted, and the following facts are discovered: 37% of the athletes are football players, 30% are basketball players, and 5% 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)
1. **State the problem:** We want to find the probability that a randomly chosen athlete plays either football or basketball.
2. **Recall the formula for the union of two events:**
$$P(F \cup B) = P(F) + P(B) - P(F \cap B)$$
where $P(F)$ is the probability of being a football player, $P(B)$ is the probability of being a basketball player, and $P(F \cap B)$ is the probability of playing both sports.
3. **Substitute the given values:**
$$P(F) = 37\% = 0.37$$
$$P(B) = 30\% = 0.30$$
$$P(F \cap B) = 5\% = 0.05$$
4. **Calculate the union probability:**
$$P(F \cup B) = 0.37 + 0.30 - 0.05$$
$$P(F \cup B) = 0.62$$
5. **Convert the probability back to a percentage:**
$$0.62 \times 100 = 62\%$$
**Final answer:** The probability that the athlete plays either football or basketball is **62%**.