1. **Stating the problem:** Given probabilities: - $P(A \cap B) = 1$ - $P(A \cap B) = 0$ - $P(A \cup B) = 1$ - $P(A) = P(B)$ - $P(A') = P(B)$ - $P(A \cup B) = P(A)$ We need to analyze these conditions and understand their implications. 2. **Analyzing the given probabilities:** - $P(A \cap B)$ cannot be both 1 and 0 simultaneously. This is a contradiction. - $P(A \cup B) = 1$ means either $A$ or $B$ or both occur with probability 1. - $P(A) = P(B)$ means the probabilities of $A$ and $B$ are equal. - $P(A') = P(B)$ means the probability of the complement of $A$ equals the probability of $B$. - $P(A \cup B) = P(A)$ means the union of $A$ and $B$ has the same probability as $A$ alone. 3. **Using the formula for union:** $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ 4. **From $P(A \cup B) = P(A)$, substitute:** $$P(A) = P(A) + P(B) - P(A \cap B)$$ 5. **Simplify:** $$0 = P(B) - P(A \cap B)$$ $$P(B) = P(A \cap B)$$ 6. **Since $P(A) = P(B)$, then:** $$P(A) = P(A \cap B)$$ 7. **Because $P(A \cap B) \leq P(A)$ always, equality means:** $$A \subseteq B$$ 8. **From $P(A') = P(B)$ and $P(A) = P(B)$, we get:** $$P(A') = P(A)$$ 9. **Since $P(A) + P(A') = 1$, then:** $$P(A) + P(A) = 1$$ $$2P(A) = 1$$ $$P(A) = \frac{1}{2}$$ 10. **Therefore:** $$P(B) = \frac{1}{2}$$ 11. **Summary:** - $P(A) = P(B) = \frac{1}{2}$ - $A \subseteq B$ - $P(A \cup B) = P(A) = \frac{1}{2}$ - $P(A \cap B) = P(B) = \frac{1}{2}$ 12. **Note:** The initial contradictory statements $P(A \cap B) = 1$ and $P(A \cap B) = 0$ cannot both be true. We used the consistent ones. **Final answer:** $$P(A) = P(B) = P(A \cap B) = P(A \cup B) = \frac{1}{2}$$