In a class of 45 students, 28 offer chemistry and 25 offer Biology. If each student offers at least one of the…

Analyze the question: We have 45 students. Each student offers Chemistry, Biology, or both. We need the probability that a randomly chosen student offers Chemistry only.

Concept: For two sets with no student outside both: n(C ∪ B) = n(C) + n(B) − n(C ∩ B)

Solution:

1. Find the number who offer both subjects: 45 = 28 + 25 − n(C ∩ B) 45 = 53 − n(C ∩ B) n(C ∩ B) = 8

2. Find the number who offer Chemistry only: Chemistry only = 28 − 8 = 20

3. Find the probability: P(Chemistry only) = 20/45 = 4/9

Answer: The probability is 4/9, which is option B.

Tip: In "at least one" problems, always find the intersection first using n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Then subtract it from the subject total to get "only" values. Check your work: Chemistry only (20) + Biology only (17) + both (8) = 45.