1. **Problem Statement:** We want to create an example of a binomial distribution and verify its consistency property. 2. **Definition:** A binomial distribution models the number of successes in $n$ independent Bernoulli trials each with success probability $p$. The probability mass function (pmf) is: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $k=0,1,2,...,n$. 3. **Consistency Property:** If $X \sim \text{Binomial}(n, p)$ and $Y \sim \text{Binomial}(m, p)$ are independent, then $X+Y \sim \text{Binomial}(n+m, p)$. 4. **Example:** Let $n=3$, $m=2$, and $p=0.5$. Define $X \sim \text{Binomial}(3, 0.5)$ and $Y \sim \text{Binomial}(2, 0.5)$. 5. **Calculate pmf of $X$:** $$P(X=k) = \binom{3}{k} (0.5)^k (0.5)^{3-k} = \binom{3}{k} (0.5)^3$$ for $k=0,1,2,3$. 6. **Calculate pmf of $Y$:** $$P(Y=j) = \binom{2}{j} (0.5)^j (0.5)^{2-j} = \binom{2}{j} (0.5)^2$$ for $j=0,1,2$. 7. **Distribution of $X+Y$:** The pmf of $X+Y$ is the convolution: $$P(X+Y = r) = \sum_{k=\max(0,r-2)}^{\min(3,r)} P(X=k) P(Y=r-k)$$ for $r=0,1,2,3,4,5$. 8. **Calculate $P(X+Y=r)$ explicitly:** Since $P(X=k) = \binom{3}{k} (0.5)^3$ and $P(Y=r-k) = \binom{2}{r-k} (0.5)^2$, $$P(X+Y=r) = (0.5)^5 \sum_{k=\max(0,r-2)}^{\min(3,r)} \binom{3}{k} \binom{2}{r-k}$$ 9. **Use Vandermonde's identity:** $$\sum_{k} \binom{3}{k} \binom{2}{r-k} = \binom{5}{r}$$ 10. **Therefore:** $$P(X+Y=r) = \binom{5}{r} (0.5)^5$$ which is the pmf of $\text{Binomial}(5, 0.5)$. 11. **Conclusion:** The sum $X+Y$ follows a binomial distribution with parameters $n+m=5$ and $p=0.5$, proving the consistency property for this example.