1. **Stating the problem:** We have a probability table for the outcomes of a game: win, lose, and draw. The probabilities for win and lose are given as 0.3 and 0.25 respectively, and we need to complete the table by finding the probability of a draw. 2. **Completing the table:** The sum of all probabilities must be 1 because one of the outcomes must happen. $$P(\text{win}) + P(\text{lose}) + P(\text{draw}) = 1$$ Given: $$P(\text{win}) = 0.3, \quad P(\text{lose}) = 0.25$$ So, $$P(\text{draw}) = 1 - (0.3 + 0.25) = 1 - 0.55 = 0.45$$ 3. **Expected number of wins in 120 games:** The expected number of wins is calculated by multiplying the probability of winning by the total number of games played. $$\text{Expected wins} = P(\text{win}) \times \text{number of games} = 0.3 \times 120 = 36$$ 4. **Estimating D using 1 significant figure:** Given: $$D = \frac{\sqrt{1.95 \times 9.92^2}}{8.07}$$ Round each number to 1 significant figure: - $1.95 \approx 2$ - $9.92 \approx 10$ - $8.07 \approx 8$ Calculate inside the square root: $$2 \times 10^2 = 2 \times 100 = 200$$ Square root: $$\sqrt{200} = \sqrt{2 \times 100} = 10 \sqrt{2} \approx 10 \times 1.4 = 14$$ Divide by 8: $$D \approx \frac{14}{8} = 1.75$$ Rounded to 1 significant figure: $$D \approx 2$$ **Final answers:** - Probability of draw: $0.45$ - Expected wins in 120 games: $36$ - Estimated $D$ to 1 significant figure: $2$