In a certain Algebra 2 class of 28 students, 21 of them play basketball and 20 of them play baseball. There are 5 students who play neither sportWhat is the probability that a student chosen randomly from the class plays both basketball and baseball?

Answer 1

Let's denote:
- A as the set of students who play basketball (21 students)
- B as the set of students who play baseball (20 students)
- AB as the set of students who play both basketball and baseball (let x be the number of students who play both sports)

We know that:
- |A| = 21
- |B| = 20
- |AB| = x
- |A ∪ B| = |A| + |B| - |AB| = 21 + 20 - x = 41 - x
- |A ∪ B| = 28 - 5 = 23 (total number of students who play at least one sport)

Using the principle of inclusion-exclusion:
|A ∪ B| = |A| + |B| - |AB|
23 = 21 + 20 - x
23 = 41 - x
23 - 41 = -x
x = 18

So, there are 18 students who play both basketball and baseball. The probability that a student chosen randomly from the class plays both sports is therefore:
P(AB) = |AB| / |Total Students| = 18 / 28 = 9/14 ≈ 0.643

Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is approximately 0.643 or 64.3%.