1. **Problem Statement:** We have two fair spinners with numbers on their sectors. The first spinner has 4 sectors labeled 6, 3, 4, 5. The second spinner has 3 sectors labeled 3, 4, 4. We spin both and add their results. We want to find the probability of scoring a total of 10. 2. **Sample Space:** The first spinner outcomes are $\{6,3,4,5\}$ and the second spinner outcomes are $\{3,4,4\}$. Since the second spinner has repeated numbers, we consider each sector separately for total outcomes. 3. **Total Number of Outcomes:** The first spinner has 4 outcomes, the second has 3 outcomes, so total outcomes = $4 \times 3 = 12$. 4. **Possible Totals:** We list all sums: - $6+3=9$ - $6+4=10$ - $6+4=10$ - $3+3=6$ - $3+4=7$ - $3+4=7$ - $4+3=7$ - $4+4=8$ - $4+4=8$ - $5+3=8$ - $5+4=9$ - $5+4=9$ 5. **Count Outcomes with Total 10:** From above, total 10 occurs twice: $(6+4)$ and $(6+4)$ (two different sectors on second spinner both labeled 4). 6. **Probability Calculation:** Probability of total 10 = $\frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{12} = \frac{1}{6}$. **Final answer:** The probability of scoring a total of 10 is $\frac{1}{6}$.