Algebra 2 class of 21 students, 6 of them play basketball and 10 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?
According to the Venn diagram that I made, where I called the intersection x
(6-x) + x + (10-x) + 7 = 21
x = 2
so prob(play both) = 2/21
2/21
In a certain Algebra 2 class of 28 students, 9 of them play basketball and 17 of them play baseball. There are 6 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball
Well, let's do some math tricks with the class of 21 students. We know that there are 6 students who play basketball, 10 students who play baseball, and 7 students who play neither sport.
Now, if we add up the number of students who play basketball and baseball, we get 6 + 10 = 16 students.
But uh-oh, we've overcounted! We know that there are only 21 students in total. So, we need to subtract the number of students who play both sports from our total count of 16.
Hmmm, since we don't know exactly how many students play both sports, we can call that number "x".
So, the equation becomes: 6 + 10 - x = 21 - 7.
Simplifying, we get: 16 - x = 14.
Now, let's isolate "x" to find out how many students play both sports: x = 16 - 14 = 2.
Aha! So, we have found out that there are 2 students who play both basketball and baseball.
Now, let's calculate the probability of choosing a student who plays both sports. We know that there are 21 students in total, so the probability is 2 (students who play both sports) divided by 21 (total students).
Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is 2/21.
To find the probability that a student plays both basketball and baseball, we need to determine how many students play both sports and divide it by the total number of students in the class.
We are given that there are 21 students in the class, 6 of whom play basketball and 10 of whom play baseball. We also know that 7 students play neither sport.
To find the number of students who play both sports, we can use the principle of inclusion-exclusion:
Number of students who play both sports = Total students who play at least one sport - Number of students who play neither sport
Total students who play at least one sport = Number of students who play basketball + Number of students who play baseball
Total students who play at least one sport = 6 + 10 = 16
Number of students who play both sports = Total students who play at least one sport - Number of students who play neither sport
Number of students who play both sports = 16 - 7 = 9
Therefore, there are 9 students who play both basketball and baseball.
Now, to find the probability, we divide the number of students who play both sports by the total number of students in the class:
Probability = Number of students who play both sports / Total number of students
Probability = 9 / 21
Simplifying the fraction, we get:
Probability = 3 / 7
So the probability that a student chosen randomly from the class plays both basketball and baseball is 3/7.