Final answer:
The probability that Clara wins exactly one of the two games is 36/121, which is calculated by using the probabilities of winning and losing each game independently and considering the two possible scenarios where she wins one game and loses the other.
Step-by-step explanation:
The student asks about the probability of Clara winning exactly one of two games. To calculate this, we need to consider two scenarios: Clara wins the first game and loses the second, and Clara loses the first game and wins the second. Assuming the probability of winning or losing each game is independent, we can use the provided probabilities.
For Game 1, the probability of winning (P(Win1)) is 9/11 and the probability of losing (P(Lose1)) is 2/11. For Game 2, the probability of winning (P(Win2)) is 9/11 and the probability of losing (P(Lose2)) is 2/11.
The probability that Clara wins the first game and loses the second is P(Win1) × P(Lose2), and the probability that Clara loses the first game and wins the second is P(Lose1) × P(Win2).
Therefore, the total probability that Clara wins exactly one game is:
P(Win1) × P(Lose2) + P(Lose1) × P(Win2) =(9/11) × (2/11) + (2/11) × (9/11) = (18/121) + (18/121) = 36/121
This fraction is already in its simplest form.