1. **State the problem:** We have 11 students in total. The Chess Club members are Aldo, Latoya, Yolanda, Melissa (4 students). The Science Club members are Sam, Goran, Latoya, Ann, Yolanda, Leila, Melissa, Lashonda, Deandre (9 students). Joe is outside both clubs. 2. **Define events:** - $A$: student is in Chess Club - $B$: student is in Science Club 3. **Calculate probabilities:** - $P(A) = \frac{\text{number in Chess}}{\text{total}} = \frac{4}{11}$ - $P(B) = \frac{\text{number in Science}}{\text{total}} = \frac{9}{11}$ - $P(A \text{ and } B) = \frac{\text{number in both clubs}}{\text{total}} = \frac{3}{11}$ (Latoya, Yolanda, Melissa) 4. **Conditional probability formula:** $$P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$$ Calculate: $$P(A|B) = \frac{\frac{3}{11}}{\frac{9}{11}} = \frac{3}{11} \times \frac{11}{9} = \frac{3}{9} = \frac{1}{3}$$ 5. **Calculate $P(B) \cdot P(A|B)$:** $$P(B) \cdot P(A|B) = \frac{9}{11} \times \frac{1}{3} = \frac{9}{33} = \frac{3}{11}$$ 6. **Answer for part (a):** - $P(A) = \frac{4}{11}$ - $P(B) = \frac{9}{11}$ - $P(A \text{ and } B) = \frac{3}{11}$ - $P(A|B) = \frac{1}{3}$ - $P(B) \cdot P(A|B) = \frac{3}{11}$ 7. **Answer for part (b):** The probability equal to $P(A \text{ and } B)$ is $\frac{3}{11}$. This matches $P(B) \cdot P(A|B)$, confirming the multiplication rule for conditional probability.