1. **Stating the problem:** We have two events $A$ (Chess club members) and $B$ (Science club members) with given probabilities: $$P(A) = \frac{9}{13}, \quad P(B) = \frac{5}{13}, \quad P(A \text{ and } B) = \frac{4}{13}$$ We need to find $P(A|B)$ (the probability of $A$ given $B$) and verify the multiplication rule $P(B) \cdot P(A|B) = P(A \text{ and } B)$. 2. **Formula used:** The conditional probability formula is $$P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$$ This formula tells us the probability of $A$ occurring given that $B$ has occurred. 3. **Calculate $P(A|B)$:** Substitute the known values: $$P(A|B) = \frac{\frac{4}{13}}{\frac{5}{13}} = \frac{4}{13} \times \frac{13}{5} = \frac{4}{5}$$ 4. **Verify multiplication rule:** Calculate $P(B) \cdot P(A|B)$: $$P(B) \cdot P(A|B) = \frac{5}{13} \times \frac{4}{5} = \frac{4}{13}$$ This matches $P(A \text{ and } B)$, confirming the multiplication rule. 5. **Answer to the multiple choice:** The probability equal to $P(A \text{ and } B)$ is $$P(B) \cdot P(A|B)$$ **Summary:** - $P(A|B) = \frac{4}{5}$ - $P(B) \cdot P(A|B) = \frac{4}{13} = P(A \text{ and } B)$ This shows the relationship between joint and conditional probabilities clearly.