A fair $12$-sided die, with faces numbered $1$ through $12$, is rolled once. Each outcome is equally likely. What is the probability that the number rolled is **either a multiple of $5$ or an odd number** (or both)?

Use the inclusion-exclusion rule: $P(M \text{ or } O) = P(M) + P(O) - P(M \text{ and } O)$ where $M$ = "multiple of 5" and $O$ = "odd." - $M = \{5, 10\}$: $P(M) = \dfrac{2}{12}$. - $O = \{1, 3, 5, 7, 9, 11\}$: $P(O) = \dfrac{6}{12}$. - $M \cap O = \{5\}$: $P(M \text{ and } O) = \dfrac{1}{12}$. $P(M \text{ or } O) = \dfrac{2}{12} + \dfrac{6}{12} - \dfrac{1}{12} = \dfrac{7}{12}$. Check by direct count: the favorable outcomes are $\{1, 3, 5, 7, 9, 10, 11\}$ — exactly $7$ of the $12$ equally-likely numbers. ✓ - Trap A ($6/12$) is just $P(O)$ alone, forgetting the $10$. - Trap C ($8/12$) adds without removing the overlap at 5 — that's the classic over-count. - Trap D ($2/3 = 8/12$) is the same overcount expressed as a fraction.