1. **Problem Statement:** We have two unbiased dice with random variables $X$ and $Y$ representing the numbers on each die. We want to find the probability mass functions (PMFs) of $X+Y$ and $X-Y$, and compute their expected values. 2. **PMF of $X+Y$:** Since $X$ and $Y$ are independent and uniformly distributed on $\{1,2,3,4,5,6\}$, the sum $S = X+Y$ ranges from 2 to 12. The PMF is given by: $$P(S = k) = \frac{\text{number of pairs }(x,y) \text{ with } x+y=k}{36}$$ The counts for each sum $k$ are: - $k=2$: 1 way (1+1) - $k=3$: 2 ways (1+2, 2+1) - $k=4$: 3 ways - $k=5$: 4 ways - $k=6$: 5 ways - $k=7$: 6 ways - $k=8$: 5 ways - $k=9$: 4 ways - $k=10$: 3 ways - $k=11$: 2 ways - $k=12$: 1 way 3. **PMF of $X-Y$:** The difference $D = X - Y$ ranges from $-5$ to $5$. The PMF is: $$P(D = d) = \frac{\text{number of pairs }(x,y) \text{ with } x-y=d}{36}$$ For each $d$: - $d=-5$: 1 way (1-6) - $d=-4$: 2 ways - $d=-3$: 3 ways - $d=-2$: 4 ways - $d=-1$: 5 ways - $d=0$: 6 ways - $d=1$: 5 ways - $d=2$: 4 ways - $d=3$: 3 ways - $d=4$: 2 ways - $d=5$: 1 way 4. **Expected Value Formula:** For a discrete random variable $Z$ with PMF $p(z)$, $$E(Z) = \sum_z z \cdot p(z)$$ 5. **Calculate $E(X+Y)$:** Since $X$ and $Y$ are independent and identically distributed, $$E(X+Y) = E(X) + E(Y)$$ Each die has expected value: $$E(X) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5$$ Therefore, $$E(X+Y) = 3.5 + 3.5 = 7$$ 6. **Calculate $E(X-Y)$:** Since $X$ and $Y$ are independent, $$E(X-Y) = E(X) - E(Y) = 3.5 - 3.5 = 0$$ **Final answers:** - PMF of $X+Y$ is as described with sums 2 to 12 and probabilities proportional to counts over 36. - PMF of $X-Y$ is as described with differences from -5 to 5 and probabilities proportional to counts over 36. - $E(X+Y) = 7$ - $E(X-Y) = 0$