1. **State the problem:** We want to find the probability that Clara wins exactly one of the two games. 2. **Identify probabilities from the tree:** - Probability of winning Game 1: $\frac{9}{11}$ - Probability of losing Game 1: $\frac{2}{11}$ - Probability of winning Game 2: $\frac{9}{11}$ - Probability of losing Game 2: $\frac{2}{11}$ 3. **Formula for exactly one win:** The event "exactly one win" means either: - Win Game 1 and Lose Game 2, or - Lose Game 1 and Win Game 2. So, $$P(\text{exactly one win}) = P(W_1 \cap L_2) + P(L_1 \cap W_2)$$ 4. **Calculate each probability:** - $P(W_1 \cap L_2) = P(W_1) \times P(L_2) = \frac{9}{11} \times \frac{2}{11} = \frac{18}{121}$ - $P(L_1 \cap W_2) = P(L_1) \times P(W_2) = \frac{2}{11} \times \frac{9}{11} = \frac{18}{121}$ 5. **Add the probabilities:** $$P(\text{exactly one win}) = \frac{18}{121} + \frac{18}{121} = \frac{36}{121}$$ 6. **Simplify the fraction:** $$\frac{36}{121} = \frac{\cancel{36}}{\cancel{121}} \text{ (no common factors to cancel)}$$ So the probability is $\frac{36}{121}$. **Final answer:** $$\boxed{\frac{36}{121}}$$