1. **State the problem:** We have a binomial distribution with mean $\mu = 6$ and standard deviation $\sigma = \sqrt{2}$. We need to find the first two terms of the distribution, i.e., $P(X=0)$ and $P(X=1)$. 2. **Recall formulas:** For a binomial distribution $X \sim Bin(n,p)$, the mean and variance are given by: $$\mu = np$$ $$\sigma^2 = np(1-p)$$ where $n$ is the number of trials and $p$ is the probability of success in each trial. 3. **Use given values:** Given $\mu = 6$ and $\sigma = \sqrt{2}$, so variance $\sigma^2 = 2$. From the formulas: $$np = 6$$ $$np(1-p) = 2$$ 4. **Find $p$ and $n$:** From $np = 6$, we get $n = \frac{6}{p}$. Substitute into variance equation: $$6(1-p) = 2$$ $$6 - 6p = 2$$ $$6p = 4$$ $$p = \frac{2}{3}$$ Then, $$n = \frac{6}{2/3} = 6 \times \frac{3}{2} = 9$$ 5. **Calculate first two terms:** The binomial probability mass function is: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ - For $k=0$: $$P(X=0) = \binom{9}{0} \left(\frac{2}{3}\right)^0 \left(\frac{1}{3}\right)^9 = 1 \times 1 \times \left(\frac{1}{3}\right)^9 = \left(\frac{1}{3}\right)^9$$ - For $k=1$: $$P(X=1) = \binom{9}{1} \left(\frac{2}{3}\right)^1 \left(\frac{1}{3}\right)^8 = 9 \times \frac{2}{3} \times \left(\frac{1}{3}\right)^8 = 9 \times \frac{2}{3} \times \frac{1}{3^8} = 6 \times \frac{1}{3^8}$$ 6. **Final answers:** $$P(X=0) = \left(\frac{1}{3}\right)^9$$ $$P(X=1) = 6 \times \left(\frac{1}{3}\right)^8$$