Question

In a certain Algebra 2 class of 21 students, 10 of them play basketball and 9 of them

play baseball. There are 8 students who play neither sport. What is the probability
that a student chosen randomly from the class plays both basketball and baseball?

Answer 1

Explanation:

To find the probability that a student plays both basketball and baseball, we need to determine the number of students who play both sports and divide it by the total number of students in the class.

Given:

Total number of students (n) = 21

Number of students who play basketball (B) = 10

Number of students who play baseball (A) = 9

Number of students who play neither sport = 8

Let's calculate the number of students who play both basketball and baseball (B ∩ A):

Number of students who play both sports (B ∩ A) = Number of students who play basketball (B) + Number of students who play baseball (A) - Total number of students (n) + Number of students who play neither sport

B ∩ A = B + A - n + Neither

B ∩ A = 10 + 9 - 21 + 8

B ∩ A = 6

The number of students who play both basketball and baseball is 6.

Now, we can calculate the probability:

Probability of playing both basketball and baseball = Number of students who play both sports (B ∩ A) / Total number of students (n)

Probability = 6 / 21

Probability = 2 / 7

Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is 2/7.