1. **Problem statement:** Given two events A and B with probabilities $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, and $P(AB) = \frac{1}{6}$, find the following probabilities: 2. **Recall important formulas:** - $P(A \cup B) = P(A) + P(B) - P(AB)$ - $P(\bar{A}) = 1 - P(A)$ - $P(\bar{B}) = 1 - P(B)$ - $P(A|B) = \frac{P(AB)}{P(B)}$ - $P(AB^c) = P(A) - P(AB)$ - $P(\bar{A}B) = P(B) - P(AB)$ - $P(\bar{A} \cup \bar{B}) = 1 - P(A \cap B)$ 3. **Calculate each part:** 1) $P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$ 2) $P(\bar{A} \cup \bar{B}) = 1 - P(A \cap B) = 1 - \frac{1}{6} = \frac{5}{6}$ 3) $P(\bar{A} \cup B) = 1 - P(A \cap \bar{B})$ Calculate $P(A \cap \bar{B}) = P(A) - P(AB) = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$ So, $P(\bar{A} \cup B) = 1 - \frac{1}{3} = \frac{2}{3}$ 4) $P(AB^c) = P(A) - P(AB) = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ 5) $P(A B^c)$ is the same as $P(AB^c) = \frac{1}{3}$ 6) $P(\bar{A} B) = P(B) - P(AB) = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$ 7) $P(\bar{A} \cup B)$ was calculated in (3) as $\frac{2}{3}$ 8) $P(A|B) = \frac{P(AB)}{P(B)} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2}$ 9) $P(A|B^c) = \frac{P(AB^c)}{P(B^c)}$ Calculate $P(B^c) = 1 - \frac{1}{3} = \frac{2}{3}$ So, $P(A|B^c) = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$ 10) $P(AB|B) = \frac{P(AB)}{P(B)} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2}$ 11) $P(AB^c|B) = \frac{P(AB^c \cap B)}{P(B)} = 0$ because $AB^c$ and $B$ are disjoint 12) $P(AB^c|B^c) = \frac{P(AB^c)}{P(B^c)} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$ 13) $P(A \cup B|AB^c) = 1$ because $AB^c$ is a subset of $A \cup B$ 14) $P(AB^c|\bar{A} \cup B) = \frac{P(AB^c \cap (\bar{A} \cup B))}{P(\bar{A} \cup B)}$ Note $AB^c \cap \bar{A} = \emptyset$ and $AB^c \cap B = \emptyset$, so $P(AB^c \cap (\bar{A} \cup B)) = 0$ Therefore, $P(AB^c|\bar{A} \cup B) = 0$ **Final answers:** 1) $\frac{2}{3}$ 2) $\frac{5}{6}$ 3) $\frac{2}{3}$ 4) $\frac{1}{3}$ 5) $\frac{1}{3}$ 6) $\frac{1}{6}$ 7) $\frac{2}{3}$ 8) $\frac{1}{2}$ 9) $\frac{1}{2}$ 10) $\frac{1}{2}$ 11) $0$ 12) $\frac{1}{2}$ 13) $1$ 14) $0$