Conditional Probability Dice E047Ad

1. **Problem statement:** Two dice are rolled. Define events: - $A$: sum of two dice equals $3$ - $B$: sum of two dice equals $7$ - $C$: at least one die shows a $1$ 2. **Total sample space:** When rolling two dice, total outcomes = $6 \times 6 = 36$. 3. **Event $C$ (at least one die shows a 1):** Possible outcomes for $C$ are pairs where first or second die is $1$: $$C = \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(3,1),(4,1),(5,1),(6,1)\}$$ Number of outcomes in $C$ is $11$. 4. **Event $A$ (sum = 3):** Possible pairs: $$A = \{(1,2),(2,1)\}$$ Number of outcomes in $A$ is $2$. 5. **Event $B$ (sum = 7):** Possible pairs: $$B = \{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$$ Number of outcomes in $B$ is $6$. 6. **Calculate $P(A \mid C)$:** By definition, $$P(A \mid C) = \frac{P(A \cap C)}{P(C)}$$ - Find $A \cap C$: outcomes in both $A$ and $C$. - $A = \{(1,2),(2,1)\}$ and $C$ includes these pairs if at least one die is $1$. - Both $(1,2)$ and $(2,1)$ have a $1$, so $A \cap C = \{(1,2),(2,1)\}$. - Number of outcomes in $A \cap C$ is $2$. Calculate probabilities: $$P(C) = \frac{11}{36}$$ $$P(A \cap C) = \frac{2}{36}$$ So, $$P(A \mid C) = \frac{\frac{2}{36}}{\frac{11}{36}} = \frac{2}{11}$$ 7. **Calculate $P(B \mid C)$:** Similarly, $$P(B \mid C) = \frac{P(B \cap C)}{P(C)}$$ - Find $B \cap C$: outcomes in both $B$ and $C$. - $B = \{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$ - From these, $(1,6)$ and $(6,1)$ have a $1$, so $$B \cap C = \{(1,6),(6,1)\}$$ - Number of outcomes in $B \cap C$ is $2$. Calculate probabilities: $$P(B \cap C) = \frac{2}{36}$$ So, $$P(B \mid C) = \frac{\frac{2}{36}}{\frac{11}{36}} = \frac{2}{11}$$ 8. **Check independence of $A$ and $C$:** Two events $A$ and $C$ are independent if $$P(A \cap C) = P(A) \times P(C)$$ Calculate $P(A)$: $$P(A) = \frac{2}{36} = \frac{1}{18}$$ Calculate $P(A) \times P(C)$: $$\frac{1}{18} \times \frac{11}{36} = \frac{11}{648}$$ Compare with $P(A \cap C) = \frac{2}{36} = \frac{1}{18} = \frac{36}{648}$ Since $\frac{36}{648} \neq \frac{11}{648}$, $A$ and $C$ are **not independent**. 9. **Check independence of $B$ and $C$:** Calculate $P(B)$: $$P(B) = \frac{6}{36} = \frac{1}{6}$$ Calculate $P(B) \times P(C)$: $$\frac{1}{6} \times \frac{11}{36} = \frac{11}{216}$$ Compare with $P(B \cap C) = \frac{2}{36} = \frac{1}{18} = \frac{12}{216}$ Since $\frac{12}{216} \neq \frac{11}{216}$, $B$ and $C$ are **not independent**. **Final answers:** - (a) $P(A \mid C) = \frac{2}{11}$ - (b) $P(B \mid C) = \frac{2}{11}$ - (c) Neither $A$ and $C$ nor $B$ and $C$ are independent.