1. The problem is to find the probability of the union of two events $A$ and $B$, given by the formula: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ 2. This formula accounts for the fact that when adding $P(A)$ and $P(B)$, the intersection $P(A \cap B)$ is counted twice, so we subtract it once. 3. Given values: $$P(A) = \frac{9}{36}, \quad P(B) = \frac{11}{36}, \quad P(A \cap B) = \frac{3}{36}$$ 4. Substitute these into the formula: $$P(A \cup B) = \frac{9}{36} + \frac{11}{36} - \frac{3}{36}$$ 5. Simplify the right side: $$P(A \cup B) = \frac{9 + 11 - 3}{36} = \frac{17}{36}$$ 6. Therefore, the probability of $A$ or $B$ occurring is $\frac{17}{36}$. 7. The other probabilities given are $\frac{1}{4}$ and $\frac{11}{36}$, which likely represent $P(A \cap B^c)$ or other related probabilities, but the main question was about $P(A \cup B)$. Final answer: $$P(A \cup B) = \frac{17}{36}$$