1. **State the problem:** We roll a fair dice twice. Event R: sum of outcomes is 7. Event S: product of outcomes is 12. Find $P(R)$, $P(S)$, $P(R \cap S)$ and determine if R and S are independent. 2. **Total sample space:** Since each dice has 6 sides, total outcomes are $6 \times 6 = 36$. 3. **Event R: sum is 7.** Possible pairs $(x,y)$ with $x+y=7$ are: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$. Number of such outcomes = 6. So, $P(R) = \frac{6}{36} = \frac{1}{6}$. 4. **Event S: product is 12.** Possible pairs $(x,y)$ such that $x \times y = 12$ and $x,y \in \{1,...,6\}$: $(2,6), (3,4), (4,3), (6,2)$. Number of such outcomes = 4. So, $P(S) = \frac{4}{36} = \frac{1}{9}$. 5. **Intersection $R \cap S$: sum is 7 and product is 12.** Common pairs from both sets are: Event R: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ Event S: $(2,6), (3,4), (4,3), (6,2)$ Intersection: $(3,4), (4,3)$ Number of such outcomes = 2. So, $P(R \cap S) = \frac{2}{36} = \frac{1}{18}$. 6. **Check independence:** Events R and S are independent if and only if $$P(R \cap S) = P(R) \times P(S)$$ Calculate the right side: $$P(R) \times P(S) = \frac{1}{6} \times \frac{1}{9} = \frac{1}{54}$$ Since $$P(R \cap S) = \frac{1}{18} \neq \frac{1}{54} = P(R) \times P(S)$$ Therefore, R and S are **not independent**. **Final answers:** $$P(R) = \frac{1}{6}, \quad P(S) = \frac{1}{9}, \quad P(R \cap S) = \frac{1}{18}$$ Events R and S are not independent.