1. **Problem Statement:** Two fair dice are thrown simultaneously. Let the random variable $X$ be the sum of the two outcomes. We need to find the probabilities for each possible sum, construct a probability table, and sketch the graph of $X$. 2. **Possible sums:** The smallest sum is $2$ (1+1) and the largest sum is $12$ (6+6). So, $X$ can take values from $2$ to $12$. 3. **Total outcomes:** Since each die has 6 faces, total outcomes are $6 \times 6 = 36$. 4. **Calculate probabilities:** The number of ways to get each sum is: - $2$: 1 way (1+1) - $3$: 2 ways (1+2, 2+1) - $4$: 3 ways (1+3, 2+2, 3+1) - $5$: 4 ways (1+4, 2+3, 3+2, 4+1) - $6$: 5 ways (1+5, 2+4, 3+3, 4+2, 5+1) - $7$: 6 ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1) - $8$: 5 ways (2+6, 3+5, 4+4, 5+3, 6+2) - $9$: 4 ways (3+6, 4+5, 5+4, 6+3) - $10$: 3 ways (4+6, 5+5, 6+4) - $11$: 2 ways (5+6, 6+5) - $12$: 1 way (6+6) Probability for each sum $x$ is $P(X=x) = \frac{\text{number of ways}}{36}$. 5. **Probability table:** | $x$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | |-----|---|---|---|---|---|---|---|---|----|----|----| | $P(X=x)$ | $\frac{1}{36}$ | $\frac{2}{36}$ | $\frac{3}{36}$ | $\frac{4}{36}$ | $\frac{5}{36}$ | $\frac{6}{36}$ | $\frac{5}{36}$ | $\frac{4}{36}$ | $\frac{3}{36}$ | $\frac{2}{36}$ | $\frac{1}{36}$ | 6. **Check sum of probabilities:** $$\sum_{x=2}^{12} P(X=x) = \frac{1+2+3+4+5+6+5+4+3+2+1}{36} = \frac{36}{36} = 1$$ 7. **Graph sketch:** The graph of $P(X=x)$ vs $x$ is a symmetric distribution with peak at $x=7$. **Final answer:** The probabilities for sums 2 to 12 are as listed in the table above.