Question:

Step 1: Identify the multiples of 2 and 3 on a single die. On a standard die showing faces $\{1, 2, 3, 4, 5, 6\}$: - Multiples of 2: $\{2, 4, 6\}$ with probability $P(\text{multiple of 2}) = \dfrac{3}{6} = \dfrac{1}{2}$ - Multiples of 3: $\{3, 6\}$ with probability $P(\text{multiple of 3}) = \dfrac{2}{6} = \dfrac{1}{3}$ Step 2: Define the two mutually exclusive favorable events. When two dice are thrown together, we want either: - Event $A$: First die shows a multiple of 2 AND second die shows a multiple of 3 - Event $B$: First die shows a multiple of 3 AND second die shows a multiple of 2 These events can overlap, so we need to use the inclusion-exclusion principle. Step 3: Calculate the probability of Event A. $$P(A) = P(\text{die 1 is multiple of 2}) \times P(\text{die 2 is multiple of 3})$$ $$P(A) = \dfrac{3}{6} \times \dfrac{2}{6} = \dfrac{6}{36}$$ Step 4: Calculate the probability of Event B. $$P(B) = P(\text{die 1 is multiple of 3}) \times P(\text{die 2 is multiple of 2})$$ $$P(B) = \dfrac{2}{6} \times \dfrac{3}{6} = \dfrac{6}{36}$$ Step 5: Calculate the probability of the intersection $A \cap B$. The intersection occurs when both conditions are satisfied simultaneously on both dice. This happens only when both dice show 6 (since 6 is the only number that is both a multiple of 2 and a multiple of 3): $$P(A \cap B) = P(\text{die 1 shows 6}) \times P(\text{die 2 shows 6})$$ $$P(A \cap B) = \dfrac{1}{6} \times \dfrac{1}{6} = \dfrac{1}{36}$$ Step 6: Apply the inclusion-exclusion principle to find the total probability. $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ $$P(A \cup B) = \dfrac{6}{36} + \dfrac{6}{36} - \dfrac{1}{36} = \dfrac{11}{36}$$ **Final Answer:** The probability is $\dfrac{11}{36}$, which corresponds to **Option 4**.