1. **State the problem:** Given probabilities $P(A) = \frac{1}{2}$, $P(B) = \frac{2}{3}$, $P(A \cap B) = \frac{1}{4}$, and $P(A \cup B) = \frac{1}{6}$, determine if events $A$ and $B$ are mutually exclusive. 2. **Recall the formula for union of two events:** $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ 3. **Substitute the given values:** $$P(A \cup B) = \frac{1}{2} + \frac{2}{3} - \frac{1}{4}$$ 4. **Find a common denominator and simplify:** $$\frac{1}{2} = \frac{6}{12}, \quad \frac{2}{3} = \frac{8}{12}, \quad \frac{1}{4} = \frac{3}{12}$$ So, $$P(A \cup B) = \frac{6}{12} + \frac{8}{12} - \frac{3}{12} = \frac{11}{12}$$ 5. **Compare with given $P(A \cup B) = \frac{1}{6}$:** The calculated $P(A \cup B) = \frac{11}{12}$ does not equal the given $\frac{1}{6}$. 6. **Conclusion:** Since the given $P(A \cup B)$ is inconsistent with the other probabilities, the events cannot be mutually exclusive because if they were, $P(A \cap B)$ would be zero, which contradicts the given $P(A \cap B) = \frac{1}{4}$. **Final answer:** No, $A$ and $B$ are not mutually exclusive.