Step 1: Understanding the Problem:
This question asks for the probability of the union of the complements of two events, $P(\bar{A} \cup \bar{B})$, when two fair dice are thrown.
Step 2: Key Formula or Approach:
According to De Morgan's Laws for probability:
\[ \bar{A} \cup \bar{B} = \overline{A \cap B} \]
Using the complement rule, we can calculate the probability as:
\[ P(\bar{A} \cup \bar{B}) = 1 - P(A \cap B) \]
Step 3: Detailed Explanation:
• The total number of possible outcomes when two fair dice are rolled is:
\[ n(S) = 6 \times 6 = 36 \]
• Let us define the sets of favorable outcomes for events $A$ and $B$:
• Event $A$ (the sum of the two faces is 9):
\[ A = \{(3, 6), (4, 5), (5, 4), (6, 3)\} \]
• Event $B$ (at least one face is 6):
\[ B = \{(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)\} \]
• The intersection event $A \cap B$ represents the outcomes where the sum of the faces is 9 AND at least one face is 6:
\[ A \cap B = \{(3, 6), (6, 3)\} \]
• The number of elements in the intersection set is:
\[ n(A \cap B) = 2 \]
• The probability of this intersection event is:
\[ P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36} = \frac{1}{18} \]
• Now, we apply the complement rule to find the probability of the union of the complements:
\[ P(\bar{A} \cup \bar{B}) = 1 - P(A \cap B) \]
\[ P(\bar{A} \cup \bar{B}) = 1 - \frac{1}{18} = \frac{17}{18} \]
Step 4: Final Answer
Thus, the probability $P(\bar{A} \cup \bar{B})$ is $\frac{17}{18}$, which corresponds to option (A).