1. **State the problem:** We want to find the probability that Arjan chooses the same move in both games of rock-paper-scissors. 2. **Understand the probabilities:** The probability of choosing each move in Game 1 is: - Rock: $\frac{7}{10}$ - Paper: $\frac{2}{10}$ - Scissors: $\frac{1}{10}$ For Game 2, the probabilities are the same regardless of the first move: - Rock: $\frac{7}{10}$ - Paper: $\frac{2}{10}$ - Scissors: $\frac{1}{10}$ 3. **Calculate the probability of choosing the same move in both games:** We calculate the probability for each move being chosen twice and then sum these probabilities. - Probability of Rock then Rock: $$\frac{7}{10} \times \frac{7}{10} = \frac{49}{100}$$ - Probability of Paper then Paper: $$\frac{2}{10} \times \frac{2}{10} = \frac{4}{100}$$ - Probability of Scissors then Scissors: $$\frac{1}{10} \times \frac{1}{10} = \frac{1}{100}$$ 4. **Sum the probabilities:** $$\frac{49}{100} + \frac{4}{100} + \frac{1}{100} = \frac{54}{100}$$ 5. **Simplify the fraction:** $$\frac{54}{100} = \frac{\cancel{54}^{27}}{\cancel{100}^{50}}$$ So the simplified probability is: $$\frac{27}{50}$$ **Final answer:** The probability that Arjan chooses the same move in both games is $\frac{27}{50}$.