1. **Problem statement:** We analyze probabilities of events A, B, and C occurring, including cases of mutual exclusivity. 2. **(a) Probability of A and B occurring simultaneously when A and B are mutually exclusive:** If A and B are mutually exclusive, they cannot happen at the same time. Therefore, $$P(A \cap B) = 0$$. 3. **(b) Expression for probability that at least one of A, B, or C occurs (no exclusivity assumed):** Using the inclusion-exclusion principle: $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$ 4. **(c) Expression when A and B are mutually exclusive, but A and C and B and C are not:** Since A and B are mutually exclusive, $$P(A \cap B) = 0$$. So, $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - 0 - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$ 5. **(d) Expression when A and B and A and C are mutually exclusive, but B and C are not:** Here, $$P(A \cap B) = 0$$ and $$P(A \cap C) = 0$$. So, $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - 0 - 0 - P(B \cap C) + P(A \cap B \cap C)$$ 6. **(e) Expression when A, B, and C are mutually exclusive of each other:** All pairwise intersections are zero: $$P(A \cap B) = P(A \cap C) = P(B \cap C) = 0$$ Also, triple intersection is zero: $$P(A \cap B \cap C) = 0$$ Therefore, $$P(A \cup B \cup C) = P(A) + P(B) + P(C)$$ **Final answers:** (a) $$P(A \cap B) = 0$$ (b) $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$ (c) $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$$ (d) $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(B \cap C) + P(A \cap B \cap C)$$ (e) $$P(A \cup B \cup C) = P(A) + P(B) + P(C)$$