To solve this probability problem, we are asked to find the probability that Clara wins exactly one of the two games. The probabilities for each possible outcome are shown in the diagram. Let's break it down.

Given Probabilities:

• 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 = $\frac{2}{11}$

Step-by-Step Calculation:

Clara wins exactly one game if:

1. She wins Game 1 and loses Game 2.
2. She loses Game 1 and wins Game 2.

Case 1: Wins Game 1 and Loses Game 2

• Probability = $\frac{9}{11} \times \frac{2}{11} = \frac{18}{121}$

Case 2: Loses Game 1 and Wins Game 2

• Probability = $\frac{2}{11} \times \frac{9}{11} = \frac{18}{121}$

Total Probability of Winning Exactly One Game:

Add the probabilities from the two cases: $\frac{18}{121} + \frac{18}{121} = \frac{36}{121}$

Thus, the probability that Clara wins exactly one of the two games is $\frac{36}{121}$.

Would you like further details or have any other questions?

Related Questions:

1. What is the probability that Clara wins both games?
2. What is the probability that Clara loses both games?
3. How would the probability change if Clara's chance of winning the second game was different?
4. What is the complement probability of Clara not winning exactly one game?
5. How do independent events influence the calculation of combined probabilities?

Tip: Always make sure that probabilities add up to 1 for a complete set of possible outcomes in a scenario.