1. **Problem:** Find $P(Y \cap Z)$ given $P(Y \cup Z) = \frac{2}{3}$, $P(Y) = \frac{2}{9}$, and $P(Z) = \frac{1}{2}$. 2. **Formula:** Use the formula for the union of two events: $$P(Y \cup Z) = P(Y) + P(Z) - P(Y \cap Z)$$ This formula accounts for the overlap counted twice when adding $P(Y)$ and $P(Z)$. 3. **Substitute values:** $$\frac{2}{3} = \frac{2}{9} + \frac{1}{2} - P(Y \cap Z)$$ 4. **Find common denominator and simplify:** Common denominator for $\frac{2}{9}$ and $\frac{1}{2}$ is 18. $$\frac{2}{9} = \frac{4}{18}, \quad \frac{1}{2} = \frac{9}{18}$$ So, $$\frac{2}{3} = \frac{4}{18} + \frac{9}{18} - P(Y \cap Z) = \frac{13}{18} - P(Y \cap Z)$$ 5. **Convert $\frac{2}{3}$ to eighteenth denominator:** $$\frac{2}{3} = \frac{12}{18}$$ 6. **Solve for $P(Y \cap Z)$:** $$\frac{12}{18} = \frac{13}{18} - P(Y \cap Z) \implies P(Y \cap Z) = \frac{13}{18} - \frac{12}{18} = \frac{1}{18}$$ **Final answer:** $$\boxed{P(Y \cap Z) = \frac{1}{18}}$$