In a certain Algebra 2 class of 30 students, 12 of them play basketball and 14 of them play baseball. There are 7 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?Answer:
Q. In a certain Algebra 2 class of 30 students, 12 of them play basketball and 14 of them play baseball. There are 7 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?Answer:
Calculate Total Students: First, let's determine the total number of students who play at least one sport. Since there are 7 students who play neither sport, we subtract this number from the total number of students to find the number who play at least one sport.Total number of students = 30Number of students who play neither sport = 7Number of students who play at least one sport = Total number of students - Number of students who play neither sportNumber of students who play at least one sport = 30−7=23
Find Students Playing Both: Next, we need to find out how many students play both basketball and baseball. We can use the principle of inclusion-exclusion to find this number. According to this principle, the number of students who play both sports is equal to the sum of the number of basketball players and the number of baseball players minus the number of students who play at least one sport.Number of students who play basketball = 12Number of students who play baseball = 14Number of students who play both basketball and baseball = Number of students who play basketball + Number of students who play baseball - Number of students who play at least one sportNumber of students who play both basketball and baseball = 12+14−23=3
Calculate Probability: Now that we know 3 students play both basketball and baseball, we can calculate the probability that a randomly chosen student from the class plays both sports.Probability = Number of students who play both sports / Total number of studentsProbability = 303Probability = 101
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