1. **Problem Statement:** We have 14 students in Ms. Evans's class. The Tennis Club members are Felipe, Leila, Eric, Hong, Debra, Raina, Christine, Aldo, Rachel (9 total including intersection). The Soccer Club members are Heather, Melissa, Yolanda, Josh, Christine, Aldo, Rachel (7 total including intersection). Elsa is not in any club. 2. **Define Events:** - $A$: student is in Tennis Club. - $B$: student is in Soccer Club. 3. **Count members:** - $|A| = 6$ (Tennis only) + 3 (intersection) = 9 - $|B| = 4$ (Soccer only) + 3 (intersection) = 7 - Total students $= 14$ 4. **Calculate probabilities:** - $P(A) = \frac{|A|}{14} = \frac{9}{14}$ - $P(B) = \frac{|B|}{14} = \frac{7}{14} = \frac{1}{2}$ - $P(A \cap B) = \frac{3}{14}$ (intersection members) 5. **Conditional probability:** - $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{3}{14}}{\frac{9}{14}} = \frac{3}{9} = \frac{1}{3}$ 6. **Check product:** - $P(A) \cdot P(B|A) = \frac{9}{14} \times \frac{1}{3} = \frac{9}{42} = \frac{3}{14} = P(A \cap B)$ 7. **Answer for (b):** The probability equal to $P(A \cap B)$ is $P(A) \cdot P(B|A)$. **Final answers:** - $P(A) = \frac{9}{14}$ - $P(B) = \frac{1}{2}$ - $P(A \cap B) = \frac{3}{14}$ - $P(B|A) = \frac{1}{3}$ - $P(A) \cdot P(B|A) = \frac{3}{14}$ (b) The correct choice is: $P(A) \cdot P(B|A)$.