1. **Problem Statement:** Given the moment generating function (mgf) of a random variable $X$ as $$M_X(t) = \left(\frac{3}{5} + \frac{1}{3} e^t\right)^9,$$ show that $$P(\mu - 2\sigma < X < \mu + 2\sigma) = \sum_{x=1}^5 \binom{9}{x} \left(\frac{1}{3}\right)^x \left(\frac{2}{3}\right)^{9-x}.$$ 2. **Understanding the mgf and distribution:** The mgf given is of the form $$M_X(t) = \left(p + q e^t\right)^n,$$ which is the mgf of a binomial distribution with parameters $n=9$, $p=\frac{3}{5}$, and $q=1-p=\frac{2}{5}$. However, the problem's mgf has $p=\frac{3}{5}$ and $q=\frac{1}{3}$ which sum to more than 1, so we need to check carefully. Actually, the mgf is $$\left(\frac{3}{5} + \frac{1}{3} e^t\right)^9,$$ but $\frac{3}{5} + \frac{1}{3} = \frac{9}{15} + \frac{5}{15} = \frac{14}{15} \neq 1,$ so this is not a standard binomial mgf. However, the problem likely assumes a binomial distribution with $p=\frac{1}{3}$ and $q=\frac{2}{3}$ (since the sum in the probability uses these). So we interpret the mgf as $$M_X(t) = \left(\frac{2}{3} + \frac{1}{3} e^t\right)^9,$$ which matches the binomial mgf with $n=9$, $p=\frac{1}{3}$. 3. **Mean and variance of binomial:** For $X \sim Binomial(n=9, p=\frac{1}{3})$, $$\mu = np = 9 \times \frac{1}{3} = 3,$$ $$\sigma^2 = npq = 9 \times \frac{1}{3} \times \frac{2}{3} = 2,$$ so $$\sigma = \sqrt{2}.$$ 4. **Calculate the interval:** $$\mu - 2\sigma = 3 - 2\sqrt{2} \approx 3 - 2.828 = 0.172,$$ $$\mu + 2\sigma = 3 + 2\sqrt{2} \approx 3 + 2.828 = 5.828.$$ 5. **Since $X$ is discrete, the probability $P(\mu - 2\sigma < X < \mu + 2\sigma)$ is:** $$P(1 \leq X \leq 5) = \sum_{x=1}^5 P(X=x) = \sum_{x=1}^5 \binom{9}{x} \left(\frac{1}{3}\right)^x \left(\frac{2}{3}\right)^{9-x}.$$ 6. **This matches the expression to be shown, completing the proof.**