- (a) The number of students who read more than one magazine is the sum of those in the intersections: 2 (A and B only) + 3 (B and C only) = 5. The total number of students is 4 + 2 + 5 + 3 + 6 + 10 = 30. Therefore, the probability is 5/30 = 1/6.
- (b) The number of students who read A or B (or both) is 4 (A only) + 2 (A and B) + 5 (B only) + 3 (B and C) = 14. The total number of students is 30. Therefore, the probability is 14/30 = 7/15.
- (c) The number of students who read both A and C is 0, as there is no region in the Venn diagram representing the intersection of A and C without B. Therefore, the probability is 0.
- (d) The number of students who read at least one magazine is 30. The number of students who read C is 3 (B and C) + 6 (C only) = 9. Therefore, given that the student reads at least one magazine, the probability that the student reads C is 9/30 = 3/10.
- (e) Let P(B) be the probability of reading magazine B: (2 + 5 + 3)/30 = 10/30 = 1/3. Let P(C) be the probability of reading magazine C: (3 + 6)/30 = 9/30 = 3/10. Let P(B ∩ C) be the probability of reading both B and C: 3/30 = 1/10. Since P(B) × P(C) = (1/3) × (3/10) = 1/10, which equals P(B ∩ C), reading magazine B and reading magazine C are statistically independent.
Parent Tip: Review the logic above to help your child master the concept of using venn diagrams problems independent practice worksheet 2.
![FREE] 15 Venn Diagrams Questions And Practice Problems Worksheet ...](/images2/using-venn-diagrams-problems-independent-practice-worksheet-2/using-venn-diagrams-problems-independent-practice-worksheet-2-25.jpg)
